leetcode417

Posted 2021-12-30 asenyang

DFS思路

 1 class Solution 
 2     private int m, n;
 3     private int[][] matrix;
 4     private int[][] direction = 0, 1, 0, -1, 1, 0, -1, 0;
 5 
 6     public List<int[]> pacificAtlantic(int[][] matrix) 
 7         List<int[]> ret = new ArrayList<>();
 8         if (matrix == null || matrix.length == 0) 
 9             return ret;
10         
11 
12         m = matrix.length;
13         n = matrix[0].length;
14         this.matrix = matrix;
15         boolean[][] canReachP = new boolean[m][n];
16         boolean[][] canReachA = new boolean[m][n];
17 
18         for (int i = 0; i < m; i++) 
19             dfs(i, 0, canReachP);//从左边界（太平洋）开始，即第0列
20             dfs(i, n - 1, canReachA);//从右边界（大西洋）开始，即最后一列
21         
22         for (int i = 0; i < n; i++) 
23             dfs(0, i, canReachP);//从上边界（太平洋）开始，即第0行
24             dfs(m - 1, i, canReachA);//从下边界（大西洋）开始，即最后一行
25         
26 
27         for (int i = 0; i < m; i++) 
28             for (int j = 0; j < n; j++) 
29                 if (canReachP[i][j] && canReachA[i][j]) //同时能从太平洋边界和大西洋边界达到的点（求交集）
30                     ret.add(new int[]i, j);
31                 
32             
33         
34 
35         return ret;
36     
37 
38     private void dfs(int r, int c, boolean[][] canReach) 
39         if (canReach[r][c]) 
40             return;
41         
42         canReach[r][c] = true;
43         for (int[] d : direction) //遍历上下左右，四个方向
44             int nextR = d[0] + r;
45             int nextC = d[1] + c;
46             if (nextR < 0 || nextR >= m || nextC < 0 || nextC >= n//越界（非法）
47                     || matrix[r][c] > matrix[nextR][nextC]) //从高处（边界）流向低处（中央区域）（非法）
48 
49                 continue;
50             
51             dfs(nextR, nextC, canReach);
52         
53     
54