Number of Islands

一、问题描述

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011

Answer: 3

二、思路

深度优先算法，执行循环，若当前为‘1’，则执行递归，递归的退出条件是：

        if(i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == '0' ||grid[i][j] == 'x')

接着将当前字符置为‘x’表示已经访问过，接着分别递归标记四个方向，依次为右、左、下、上，遍历与顺序无关。

三、代码

class Solution {public:    void dfs(vector<vector<char>>& grid,int i, int j, int m, int n){        if(i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == '0' ||grid[i][j] == 'x')            return ;        grid[i][j] = 'x';        dfs(grid,i,j - 1,m , n);        dfs(grid,i,j + 1,m , n);        dfs(grid,i + 1,j,m , n);        dfs(grid,i - 1,j,m , n);    }    int numIslands(vector<vector<char>>& grid) {        int m = grid.size();        if(m < 1) return 0;        int n = grid[0].size();        int count = 0;        for(int i = 0; i < m; ++i){            for(int j = 0; j < n; ++j){                if(grid[i][j] == '1'){                    dfs(grid,i,j,m,n);                    ++count;                }             }        }        return count;    }};