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feat: add solutions to lc problem: No.0864
No.0864.Shortest Path to Get All Keys
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Diff for: ‎solution/0800-0899/0864.Shortest Path to Get All Keys/README.md

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@@ -67,22 +67,231 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：状态压缩 + BFS**
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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 shortestPathAllKeys(self, grid: List[str]) -> int:
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m, n = len(grid), len(grid[0])
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cnt, start = 0, None
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for i, row in enumerate(grid):
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for j, v in enumerate(row):
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cnt += v.islower()
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if v == '@':
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start = (i, j)
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q = deque([(start[0], start[1], 0)])
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dirs = [-1, 0, 1, 0, -1]
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ans = 0
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mask = (1 << cnt) - 1
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vis = set([(start[0], start[1], 0)])
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while q:
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for _ in range(len(q)):
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i, j, state = q.popleft()
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if state == mask:
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return ans
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for k in range(4):
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nxt = state
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x, y = i + dirs[k], j + dirs[k + 1]
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if 0 <= x < m and 0 <= y < n and grid[x][y] != '#':
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if grid[x][y].isupper() and (nxt & (1 << (ord(grid[x][y]) - ord('A')))) == 0:
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continue
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if grid[x][y].islower():
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nxt |= 1 << (ord(grid[x][y]) - ord('a'))
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if (x, y, nxt) not in vis:
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q.append((x, y, nxt))
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vis.add((x, y, nxt))
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ans += 1
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return -1
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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 int shortestPathAllKeys(String[] grid) {
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int m = grid.length, n = grid[0].length();
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int cnt = 0;
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int sx = 0, sy = 0;
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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char c = grid[i].charAt(j);
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if (Character.isLowerCase(c)) {
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++cnt;
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} else if (c == '@') {
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sx = i;
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sy = j;
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}
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}
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}
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Deque<int[]> q = new ArrayDeque<>();
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q.offer(new int[]{sx, sy, 0});
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int[] dirs = {-1, 0, 1, 0, -1};
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int ans = 0;
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int mask = (1 << cnt) - 1;
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boolean[][][] vis = new boolean[m][n][1 << cnt];
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vis[sx][sy][0] = true;
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while (!q.isEmpty()) {
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for (int t = q.size(); t > 0; --t) {
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int[] p = q.poll();
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int i = p[0], j = p[1], state = p[2];
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if (state == mask) {
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return ans;
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}
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for (int k = 0; k < 4; ++k) {
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int nxt = state;
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int x = i + dirs[k], y = j + dirs[k + 1];
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if (x >= 0 && x < m && y >= 0 && y < n) {
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char c = grid[x].charAt(y);
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if (c == '#' || (Character.isUpperCase(c) && (nxt & (1 << (c - 'A'))) == 0)) {
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continue;
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}
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if (Character.isLowerCase(c)) {
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nxt |= 1 << (c - 'a');
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}
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if (!vis[x][y][nxt]) {
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vis[x][y][nxt] = true;
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q.offer(new int[]{x, y, nxt});
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}
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}
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}
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}
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++ans;
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}
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return -1;
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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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int shortestPathAllKeys(vector<string>& grid) {
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int m = grid.size(), n = grid[0].size();
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int cnt = 0;
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int sx = 0, sy = 0;
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for (int i = 0; i < m; ++i)
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{
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for (int j = 0; j < n; ++j)
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{
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char c = grid[i][j];
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if (islower(c)) ++cnt;
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else if (c == '@')
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{
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sx = i;
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sy = j;
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}
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}
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}
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queue<vector<int>> q;
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q.push({sx, sy, 0});
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int mask = (1 << cnt) - 1;
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vector<vector<vector<bool>>> vis(m, vector<vector<bool>>(n, vector<bool>(1 << cnt)));
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vis[sx][sy][0] = true;
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int ans = 0;
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vector<int> dirs = {-1, 0, 1, 0, -1};
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while (!q.empty())
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{
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for (int t = q.size(); t; --t)
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{
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auto p = q.front();
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q.pop();
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int i = p[0], j = p[1], state = p[2];
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if (state == mask) return ans;
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for (int k = 0; k < 4; ++k)
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{
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int nxt = state;
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int x = i + dirs[k], y = j + dirs[k + 1];
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if (x >= 0 && x < m && y >= 0 && y < n)
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{
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char c = grid[x][y];
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if (c == '#' || (isupper(c) && (nxt & (1 << (c - 'A'))) == 0)) continue;
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if (islower(c)) nxt |= 1 << (c - 'a');
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if (!vis[x][y][nxt])
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{
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vis[x][y][nxt] = true;
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q.push({x, y, nxt});
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}
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}
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}
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}
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++ans;
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}
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return -1;
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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 shortestPathAllKeys(grid []string) int {
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m, n := len(grid), len(grid[0])
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cnt := 0
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sx, sy := 0, 0
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for i, row := range grid {
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for j, c := range row {
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if 'a' <= c && c <= 'z' {
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cnt++
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} else if c == '@' {
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sx, sy = i, j
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}
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}
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}
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q := [][]int{{sx, sy, 0}}
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vis := make([][][]bool, m)
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for i := range vis {
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vis[i] = make([][]bool, n)
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for j := range vis[i] {
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vis[i][j] = make([]bool, 1<<cnt)
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}
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}
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vis[sx][sy][0] = true
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dirs := []int{-1, 0, 1, 0, -1}
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ans := 0
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mask := (1 << cnt) - 1
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for len(q) > 0 {
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for t := len(q); t > 0; t-- {
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p := q[0]
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q = q[1:]
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i, j, state := p[0], p[1], p[2]
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if state == mask {
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return ans
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}
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for k := 0; k < 4; k++ {
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nxt := state
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x, y := i+dirs[k], j+dirs[k+1]
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if x >= 0 && x < m && y >= 0 && y < n {
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c := grid[x][y]
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if c == '#' {
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continue
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}
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if 'A' <= c && c <= 'Z' && (nxt&(1<<(c-'A'))) == 0 {
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continue
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}
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if 'a' <= c && c <= 'z' {
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nxt |= 1 << (c - 'a')
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}
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if !vis[x][y][nxt] {
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vis[x][y][nxt] = true
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q = append(q, []int{x, y, nxt})
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}
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}
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}
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}
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ans++
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}
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return -1
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}
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```
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### **...**

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