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feat: add solutions to lcci problem: No.1622 (#1339)
No.16.22.Langtons Ant
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‎lcci/16.22.Langtons Ant/README.md

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## 题目描述
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<!-- 这里写题目描述 -->
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<p>一只蚂蚁坐在由白色和黑色方格构成的无限网格上。开始时，网格全白，蚂蚁面向右侧。每行走一步，蚂蚁执行以下操作。</p>
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<p>(1) 如果在白色方格上，则翻转方格的颜色，向右(顺时针)转 90 度，并向前移动一个单位。<br>
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(2) 如果在黑色方格上，则翻转方格的颜色，向左(逆时针方向)转 90 度，并向前移动一个单位。</p>
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<ul>
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<li><code>K &lt;= 100000</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：哈希表 + 模拟**
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我们使用哈希表 $black$ 来记录所有黑色方格的位置，哈希表 $dirs$ 来记录蚂蚁的四个方向。我们使用变量 $x, y$ 来记录蚂蚁的位置，使用变量 $p$ 来记录蚂蚁的方向。我们使用变量 $x1, y1, x2, y2$ 来记录所有黑色方格的最小横坐标、最小纵坐标、最大横坐标、最大纵坐标。
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我们模拟蚂蚁的行走过程。如果蚂蚁所在的方格是白色的，那么蚂蚁向右转 $90$ 度，将方格涂黑，向前移动一个单位。如果蚂蚁所在的方格是黑色的，那么蚂蚁向左转 $90$ 度，将方格涂白，向前移动一个单位。在模拟的过程中，我们不断更新 $x1, y1, x2, y2$ 的值，使得它们能够包含蚂蚁走过的所有方格。
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模拟结束后，我们根据 $x1, y1, x2, y2$ 的值，构造出答案矩阵 $g$。然后，我们将蚂蚁所在的位置涂上蚂蚁的方向，同时将所有黑色方格涂上 $X$，最后返回答案矩阵。
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时间复杂度 $O(K)$，空间复杂度 $O(K)$。其中 $K$ 是蚂蚁行走的步数。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def printKMoves(self, K: int) -> List[str]:
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x1 = y1 = x2 = y2 = 0
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dirs = (0, 1, 0, -1, 0)
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d = "RDLU"
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x = y = 0
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p = 0
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black = set()
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for _ in range(K):
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if (x, y) in black:
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black.remove((x, y))
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p = (p + 3) % 4
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else:
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black.add((x, y))
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p = (p + 1) % 4
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x += dirs[p]
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y += dirs[p + 1]
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x1 = min(x1, x)
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y1 = min(y1, y)
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x2 = max(x2, x)
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y2 = max(y2, y)
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m, n = x2 - x1 + 1, y2 - y1 + 1
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g = [["_"] * n for _ in range(m)]
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for i, j in black:
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g[i - x1][j - y1] = "X"
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g[x - x1][y - y1] = d[p]
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return ["".join(row) for row in g]
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```
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````
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public List<String> printKMoves(int K) {
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int x1 = 0, y1 = 0, x2 = 0, y2 = 0;
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int[] dirs = {0, 1, 0, -1, 0};
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String d = "RDLU";
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int x = 0, y = 0, p = 0;
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Set<List<Integer>> black = new HashSet<>();
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while (K-- > 0) {
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List<Integer> t = List.of(x, y);
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if (black.add(t)) {
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p = (p + 1) % 4;
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} else {
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black.remove(t);
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p = (p + 3) % 4;
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}
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x += dirs[p];
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y += dirs[p + 1];
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x1 = Math.min(x1, x);
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y1 = Math.min(y1, y);
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x2 = Math.max(x2, x);
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y2 = Math.max(y2, y);
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}
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int m = x2 - x1 + 1;
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int n = y2 - y1 + 1;
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char[][] g = new char[m][n];
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for (char[] row : g) {
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Arrays.fill(row, '_');
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}
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for (List<Integer> t : black) {
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int i = t.get(0) - x1;
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int j = t.get(1) - y1;
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g[i][j] = 'X';
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}
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g[x - x1][y - y1] = d.charAt(p);
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List<String> ans = new ArrayList<>();
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for (char[] row : g) {
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ans.add(String.valueOf(row));
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<string> printKMoves(int K) {
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int x1 = 0, y1 = 0, x2 = 0, y2 = 0;
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int dirs[5] = {0, 1, 0, -1, 0};
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string d = "RDLU";
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int x = 0, y = 0, p = 0;
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set<pair<int, int>> black;
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while (K--) {
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auto t = make_pair(x, y);
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if (black.count(t)) {
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black.erase(t);
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p = (p + 3) % 4;
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} else {
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black.insert(t);
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p = (p + 1) % 4;
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}
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x += dirs[p];
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y += dirs[p + 1];
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x1 = min(x1, x);
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y1 = min(y1, y);
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x2 = max(x2, x);
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y2 = max(y2, y);
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}
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int m = x2 - x1 + 1, n = y2 - y1 + 1;
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vector<string> g(m, string(n, '_'));
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for (auto& [i, j] : black) {
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g[i - x1][j - y1] = 'X';
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}
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g[x - x1][y - y1] = d[p];
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return g;
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}
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};
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```
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### **Go**
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```go
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func printKMoves(K int) []string {
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var x1, y1, x2, y2, x, y, p int
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dirs := [5]int{0, 1, 0, -1, 0}
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d := "RDLU"
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type pair struct{ x, y int }
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black := map[pair]bool{}
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for K > 0 {
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t := pair{x, y}
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if black[t] {
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delete(black, t)
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p = (p + 3) % 4
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} else {
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black[t] = true
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p = (p + 1) % 4
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}
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x += dirs[p]
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y += dirs[p+1]
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x1 = min(x1, x)
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y1 = min(y1, y)
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x2 = max(x2, x)
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y2 = max(y2, y)
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K--
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}
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m, n := x2-x1+1, y2-y1+1
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g := make([][]byte, m)
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for i := range g {
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g[i] = make([]byte, n)
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for j := range g[i] {
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g[i][j] = '_'
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}
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}
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for t := range black {
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i, j := t.x-x1, t.y-y1
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g[i][j] = 'X'
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}
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g[x-x1][y-y1] = d[p]
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ans := make([]string, m)
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for i := range ans {
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ans[i] = string(g[i])
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}
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return ans
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}
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````
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**
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‎lcci/16.22.Langtons Ant/README_EN.md

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### **Python3**
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```python
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class Solution:
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def printKMoves(self, K: int) -> List[str]:
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x1 = y1 = x2 = y2 = 0
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dirs = (0, 1, 0, -1, 0)
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d = "RDLU"
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x = y = 0
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p = 0
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black = set()
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for _ in range(K):
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if (x, y) in black:
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black.remove((x, y))
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p = (p + 3) % 4
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else:
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black.add((x, y))
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p = (p + 1) % 4
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x += dirs[p]
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y += dirs[p + 1]
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x1 = min(x1, x)
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y1 = min(y1, y)
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x2 = max(x2, x)
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y2 = max(y2, y)
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m, n = x2 - x1 + 1, y2 - y1 + 1
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g = [["_"] * n for _ in range(m)]
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for i, j in black:
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g[i - x1][j - y1] = "X"
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g[x - x1][y - y1] = d[p]
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return ["".join(row) for row in g]
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```
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### **Java**
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```java
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class Solution {
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public List<String> printKMoves(int K) {
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int x1 = 0, y1 = 0, x2 = 0, y2 = 0;
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int[] dirs = {0, 1, 0, -1, 0};
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String d = "RDLU";
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int x = 0, y = 0, p = 0;
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Set<List<Integer>> black = new HashSet<>();
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while (K-- > 0) {
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List<Integer> t = List.of(x, y);
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if (black.add(t)) {
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p = (p + 1) % 4;
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} else {
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black.remove(t);
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p = (p + 3) % 4;
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}
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x += dirs[p];
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y += dirs[p + 1];
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x1 = Math.min(x1, x);
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y1 = Math.min(y1, y);
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x2 = Math.max(x2, x);
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y2 = Math.max(y2, y);
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}
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int m = x2 - x1 + 1;
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int n = y2 - y1 + 1;
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char[][] g = new char[m][n];
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for (char[] row : g) {
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Arrays.fill(row, '_');
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}
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for (List<Integer> t : black) {
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int i = t.get(0) - x1;
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int j = t.get(1) - y1;
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g[i][j] = 'X';
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}
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g[x - x1][y - y1] = d.charAt(p);
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List<String> ans = new ArrayList<>();
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for (char[] row : g) {
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ans.add(String.valueOf(row));
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<string> printKMoves(int K) {
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int x1 = 0, y1 = 0, x2 = 0, y2 = 0;
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int dirs[5] = {0, 1, 0, -1, 0};
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string d = "RDLU";
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int x = 0, y = 0, p = 0;
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set<pair<int, int>> black;
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while (K--) {
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auto t = make_pair(x, y);
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if (black.count(t)) {
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black.erase(t);
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p = (p + 3) % 4;
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} else {
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black.insert(t);
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p = (p + 1) % 4;
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}
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x += dirs[p];
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y += dirs[p + 1];
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x1 = min(x1, x);
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y1 = min(y1, y);
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x2 = max(x2, x);
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y2 = max(y2, y);
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}
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int m = x2 - x1 + 1, n = y2 - y1 + 1;
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vector<string> g(m, string(n, '_'));
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for (auto& [i, j] : black) {
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g[i - x1][j - y1] = 'X';
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}
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g[x - x1][y - y1] = d[p];
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return g;
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}
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};
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```
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### **Go**
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```go
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func printKMoves(K int) []string {
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var x1, y1, x2, y2, x, y, p int
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dirs := [5]int{0, 1, 0, -1, 0}
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d := "RDLU"
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type pair struct{ x, y int }
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black := map[pair]bool{}
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for K > 0 {
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t := pair{x, y}
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if black[t] {
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delete(black, t)
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p = (p + 3) % 4
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} else {
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black[t] = true
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p = (p + 1) % 4
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}
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x += dirs[p]
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y += dirs[p+1]
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x1 = min(x1, x)
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y1 = min(y1, y)
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x2 = max(x2, x)
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y2 = max(y2, y)
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K--
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}
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m, n := x2-x1+1, y2-y1+1
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g := make([][]byte, m)
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for i := range g {
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g[i] = make([]byte, n)
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for j := range g[i] {
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g[i][j] = '_'
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}
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}
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for t := range black {
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i, j := t.x-x1, t.y-y1
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g[i][j] = 'X'
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}
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g[x-x1][y-y1] = d[p]
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ans := make([]string, m)
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for i := range ans {
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ans[i] = string(g[i])
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}
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return ans
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

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