LeetCode 200. 岛屿数量

class Solution {
    public int numIslands(char[][] grid) {
        //统计岛屿的数量
        int count = 0;
        for(int i = 0;i < grid.length;i++){
            for(int j = 0;j < grid[0].length;j++){
                if(grid[i][j] == '1'){
                    dfs(grid,i,j);
                    count++;
                }
            }
        }
        return count;
    }
    public void dfs(char[][] grid,int i,int j){
        if( i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] =='0'){
            return;
        }
        //将此点置0,避免重复搜索
        grid[i][j] = '0';
        // 从 (i, j) 向此点的上下左右 (i+1,j),(i-1,j),(i,j+1),(i,j-1) 做深度搜索
        dfs(grid,i + 1,j);
        dfs(grid,i - 1,j);
        dfs(grid,i,j + 1);
        dfs(grid,i,j - 1);   
    }
}
原文地址:https://www.cnblogs.com/peanut-zh/p/13917373.html