題目

https://leetcode.com/problems/pacific-atlantic-water-flow/

題解

一開始的錯誤思路：（區分“主動”與“被動”的過程）

正確答案參考視頻：https://leetcode-cn.com/problems/pacific-atlantic-water-flow/solution/java-si-lu-qing-xi-dai-ma-jian-ji-by-ven-4cds/

看了視頻的思路之后，發現其實這是一個經典的“感染”問題…代碼迅速搞定。

class Solution {int M;int N;public List<List<Integer>> pacificAtlantic(int[][] heights) {M = heights.length;N = heights[0].length;boolean[][] seen = new boolean[M][N];boolean[][] LU = new boolean[M][N];for (int i = 0; i < M; i++) {dfs(heights, LU, seen, i, 0);}for (int j = 0; j < N; j++) {dfs(heights, LU, seen, 0, j);}boolean[][] RD = new boolean[M][N];seen = new boolean[M][N];for (int i = M - 1; i >= 0; i--) {dfs(heights, RD, seen, i, N - 1);}for (int j = N - 1; j >= 0; j--) {dfs(heights, RD, seen, M - 1, j);}List<List<Integer>> res = new ArrayList<>();for (int i = 0; i < M; i++) {for (int j = 0; j < N; j++) {if (LU[i][j] && RD[i][j]) {ArrayList<Integer> list = new ArrayList<>();list.add(i);list.add(j);res.add(list);}}}return res;}public void dfs(int[][] m, boolean[][] A, boolean[][] seen, int i, int j) {if (i < 0 || j < 0 || i == M || j == N) return;seen[i][j] = true;A[i][j] = true;if (i > 0 && m[i - 1][j] >= m[i][j] && !seen[i - 1][j]) dfs(m, A, seen, i - 1, j);if (j > 0 && m[i][j - 1] >= m[i][j] && !seen[i][j - 1]) dfs(m, A, seen, i, j - 1);if (i < M - 1 && m[i + 1][j] >= m[i][j] && !seen[i + 1][j]) dfs(m, A, seen, i + 1, j);if (j < N - 1 && m[i][j + 1] >= m[i][j] && !seen[i][j + 1]) dfs(m, A, seen, i, j + 1);} }

總結

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