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yanglbmeyanglbme
committedFeb 5, 2022
feat: add solutions to lc problem: No.0417
No.0417.Pacific Atlantic Water Flow
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‎README.md

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@@ -26,9 +26,9 @@
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## 算法全解
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- [LeetCode](/solution/README.md)
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- [LeetCode 《剑指 Offer（第 2 版）》](/lcof/README.md)
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- [LeetCode 《剑指 Offer（专项突击版）》](/lcof2/README.md)
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- [LeetCode 《程序员面试金典（第 6 版）》](/lcci/README.md)
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- [剑指 Offer（第 2 版）](/lcof/README.md)
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- [剑指 Offer（专项突击版）](/lcof2/README.md)
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- [程序员面试金典（第 6 版）](/lcci/README.md)
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## 基础算法通关
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‎solution/0400-0499/0417.Pacific Atlantic Water Flow/README.md

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<!-- 这里可写通用的实现逻辑 -->
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反向寻找，从海洋开始逆流，只要比该陆地高的地方就能逆流。最后是寻找两个海洋的水流都能经过的陆地即为结果。
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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 pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:
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def bfs(q, vis):
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while q:
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for _ in range(len(q), 0, -1):
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i, j = q.popleft()
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for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
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x, y = i + a, j + b
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if 0 <= x < m and 0 <= y < n and (x, y) not in vis and heights[x][y] >= heights[i][j]:
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vis.add((x, y))
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q.append((x, y))
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m, n = len(heights), len(heights[0])
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vis1, vis2 = set(), set()
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q1 = deque()
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q2 = deque()
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for i in range(m):
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for j in range(n):
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if i == 0 or j == 0:
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vis1.add((i, j))
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q1.append((i, j))
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if i == m - 1 or j == n - 1:
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vis2.add((i, j))
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q2.append((i, j))
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bfs(q1, vis1)
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bfs(q2, vis2)
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ans = []
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for i in range(m):
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for j in range(n):
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if (i, j) in vis1 and (i, j) in vis2:
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ans.append((i, j))
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return ans
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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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private int[][] heights;
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private int m;
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private int n;
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public List<List<Integer>> pacificAtlantic(int[][] heights) {
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m = heights.length;
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n = heights[0].length;
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this.heights = heights;
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Deque<int[]> q1 = new LinkedList<>();
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Deque<int[]> q2 = new LinkedList<>();
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Set<Integer> vis1 = new HashSet<>();
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Set<Integer> vis2 = new HashSet<>();
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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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if (i == 0 || j == 0) {
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vis1.add(i * n + j);
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q1.offer(new int[]{i, j});
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}
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if (i == m - 1 || j == n - 1) {
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vis2.add(i * n + j);
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q2.offer(new int[]{i, j});
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}
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}
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}
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bfs(q1, vis1);
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bfs(q2, vis2);
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List<List<Integer>> ans = new ArrayList<>();
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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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int x = i * n + j;
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if (vis1.contains(x) && vis2.contains(x)) {
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ans.add(Arrays.asList(i, j));
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}
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}
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}
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return ans;
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}
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private void bfs(Deque<int[]> q, Set<Integer> vis) {
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int[] dirs = {-1, 0, 1, 0, -1};
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while (!q.isEmpty()) {
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for (int k = q.size(); k > 0; --k) {
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int[] p = q.poll();
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for (int i = 0; i < 4; ++i) {
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int x = p[0] + dirs[i];
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int y = p[1] + dirs[i + 1];
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if (x >= 0 && x < m && y >= 0 && y < n && !vis.contains(x * n + y) && heights[x][y] >= heights[p[0]][p[1]]) {
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vis.add(x * n + y);
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q.offer(new int[]{x, y});
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}
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}
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}
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}
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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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typedef pair<int, int> pii;
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class Solution {
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public:
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vector<vector<int>> heights;
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int m;
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int n;
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vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
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m = heights.size();
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n = heights[0].size();
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this->heights = heights;
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queue<pii> q1;
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queue<pii> q2;
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unordered_set<int> vis1;
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unordered_set<int> vis2;
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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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if (i == 0 || j == 0)
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{
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vis1.insert(i * n + j);
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q1.emplace(i, j);
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}
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if (i == m - 1 || j == n - 1)
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{
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vis2.insert(i * n + j);
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q2.emplace(i, j);
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}
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}
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}
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bfs(q1, vis1);
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bfs(q2, vis2);
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vector<vector<int>> ans;
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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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int x = i * n + j;
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if (vis1.count(x) && vis2.count(x))
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{
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ans.push_back({i, j});
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}
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}
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}
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return ans;
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}
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void bfs(queue<pii>& q, unordered_set<int>& vis) {
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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 k = q.size(); k > 0; --k)
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{
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auto p = q.front();
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q.pop();
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for (int i = 0; i < 4; ++i)
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{
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int x = p.first + dirs[i];
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int y = p.second + dirs[i + 1];
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if (x >= 0 && x < m && y >= 0 && y < n && !vis.count(x * n + y) && heights[x][y] >= heights[p.first][p.second])
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{
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vis.insert(x * n + y);
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q.emplace(x, y);
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}
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}
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}
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}
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}
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};
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```
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### **...**

‎solution/0400-0499/0417.Pacific Atlantic Water Flow/README_EN.md

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@@ -42,13 +42,175 @@
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### **Python3**
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```python
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class Solution:
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def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:
47+
def bfs(q, vis):
48+
while q:
49+
for _ in range(len(q), 0, -1):
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i, j = q.popleft()
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for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
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x, y = i + a, j + b
53+
if 0 <= x < m and 0 <= y < n and (x, y) not in vis and heights[x][y] >= heights[i][j]:
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vis.add((x, y))
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q.append((x, y))
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m, n = len(heights), len(heights[0])
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vis1, vis2 = set(), set()
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q1 = deque()
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q2 = deque()
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for i in range(m):
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for j in range(n):
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if i == 0 or j == 0:
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vis1.add((i, j))
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q1.append((i, j))
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if i == m - 1 or j == n - 1:
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vis2.add((i, j))
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q2.append((i, j))
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bfs(q1, vis1)
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bfs(q2, vis2)
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ans = []
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for i in range(m):
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for j in range(n):
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if (i, j) in vis1 and (i, j) in vis2:
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ans.append((i, j))
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return ans
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```
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### **Java**
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```java
82+
class Solution {
83+
private int[][] heights;
84+
private int m;
85+
private int n;
86+
87+
public List<List<Integer>> pacificAtlantic(int[][] heights) {
88+
m = heights.length;
89+
n = heights[0].length;
90+
this.heights = heights;
91+
Deque<int[]> q1 = new LinkedList<>();
92+
Deque<int[]> q2 = new LinkedList<>();
93+
Set<Integer> vis1 = new HashSet<>();
94+
Set<Integer> vis2 = new HashSet<>();
95+
for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (i == 0 || j == 0) {
98+
vis1.add(i * n + j);
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q1.offer(new int[]{i, j});
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}
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if (i == m - 1 || j == n - 1) {
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vis2.add(i * n + j);
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q2.offer(new int[]{i, j});
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}
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}
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}
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bfs(q1, vis1);
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bfs(q2, vis2);
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List<List<Integer>> ans = new ArrayList<>();
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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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int x = i * n + j;
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if (vis1.contains(x) && vis2.contains(x)) {
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ans.add(Arrays.asList(i, j));
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}
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}
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}
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return ans;
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}
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private void bfs(Deque<int[]> q, Set<Integer> vis) {
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int[] dirs = {-1, 0, 1, 0, -1};
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while (!q.isEmpty()) {
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for (int k = q.size(); k > 0; --k) {
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int[] p = q.poll();
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for (int i = 0; i < 4; ++i) {
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int x = p[0] + dirs[i];
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int y = p[1] + dirs[i + 1];
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if (x >= 0 && x < m && y >= 0 && y < n && !vis.contains(x * n + y) && heights[x][y] >= heights[p[0]][p[1]]) {
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vis.add(x * n + y);
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q.offer(new int[]{x, y});
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}
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}
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}
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}
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}
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}
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```
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### **C++**
141+
142+
```cpp
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typedef pair<int, int> pii;
144+
145+
class Solution {
146+
public:
147+
vector<vector<int>> heights;
148+
int m;
149+
int n;
150+
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vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
152+
m = heights.size();
153+
n = heights[0].size();
154+
this->heights = heights;
155+
queue<pii> q1;
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queue<pii> q2;
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unordered_set<int> vis1;
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unordered_set<int> vis2;
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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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if (i == 0 || j == 0)
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{
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vis1.insert(i * n + j);
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q1.emplace(i, j);
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}
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if (i == m - 1 || j == n - 1)
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{
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vis2.insert(i * n + j);
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q2.emplace(i, j);
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}
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}
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}
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bfs(q1, vis1);
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bfs(q2, vis2);
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vector<vector<int>> ans;
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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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int x = i * n + j;
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if (vis1.count(x) && vis2.count(x))
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{
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ans.push_back({i, j});
186+
}
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}
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}
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return ans;
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}
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void bfs(queue<pii>& q, unordered_set<int>& vis) {
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vector<int> dirs = {-1, 0, 1, 0, -1};
194+
while (!q.empty())
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{
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for (int k = q.size(); k > 0; --k)
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{
198+
auto p = q.front();
199+
q.pop();
200+
for (int i = 0; i < 4; ++i)
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{
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int x = p.first + dirs[i];
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int y = p.second + dirs[i + 1];
204+
if (x >= 0 && x < m && y >= 0 && y < n && !vis.count(x * n + y) && heights[x][y] >= heights[p.first][p.second])
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{
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vis.insert(x * n + y);
207+
q.emplace(x, y);
208+
}
209+
}
210+
}
211+
}
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}
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};
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```
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### **...**

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