Number of Islands

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本文介绍了一种使用深度优先搜索算法来统计二维网格中岛屿数量的方法。通过递归地标记相连的陆地，并将其视为一个岛屿，最终统计得到的岛屿总数。文章提供了详细的实现思路和完整的C++代码示例。

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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:

Answer: 1

Example 2:

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;
    }
};