leetcode 200 岛屿数量

思路

就是深搜，将可以走到的都标为0

##代码

class Solution {static int[][] to = {{1,0},{0,1},{-1,0},{0,-1}};public int numIslands(char[][] grid) {// 深搜int result = 0;for (int i = 0; i < grid.length; i++) {for (int j = 0; j < grid[0].length; j++) {if (grid[i][j] == '1'){dfs(i,j,grid);result++;}}}return result;}private void dfs(int i, int j, char[][] grid) {grid[i][j]='0';//标记这个点走过for (int k = 0; k < 4; k++) {int ii = i + to[k][0];int jj = j +to[k][1];if (ii >= grid.length || ii<0 ||jj>=grid[0].length|| jj < 0){continue;}if (grid[ii][jj] == '1'){dfs(ii,jj,grid);}}}
}

注意

每一个点都会循环四个方向，但是不要再让他走他来过的那个方向，要不就死循环了