LeetCode Hot 100 岛屿数量

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

示例 1：

输入：grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
输出：1

示例 2：

输入：grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
输出：3

方法一：深度优先搜索
我们可以将二维网格看成一个无向图，竖直或水平相邻的 1 之间有边相连。

为了求出岛屿的数量，我们可以扫描整个二维网格。如果一个位置为 1，则以其为起始节点开始进行深度优先搜索。在深度优先搜索的过程中，每个搜索到的 1 都会被重新标记为 0。

最终岛屿的数量就是我们进行深度优先搜索的次数。

class Solution {
public:
    void dfs(vector>&grid,int r,int c)
    {
        int nr = grid.size();
        int nc = grid[0].size();
        grid[r][c] = '0';
        if(r - 1 >=0 && grid[r - 1][c] == '1') dfs(grid,r - 1,c);
        if(r + 1 =0 && grid[r][c - 1] == '1') dfs(grid,r,c - 1);
        if(c + 1 >& grid) {
        int nr = grid.size();
        if(!nr)
        {
            return 0;
        }
        int nc = grid[0].size();
        int nums = 0;
        for(int r = 0; r < nr;r++)
        {
            for(int c = 0; c < nc;c++)
            {
                if(grid[r][c] == '1')
                {
                    nums ++;
                    dfs(grid,r,c);
                }
            }
        }
        return nums;
    }
};