2020-11-03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

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-islands
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

DFS算法：

class Solution {
private:
    void dfs(vector<vector<char>>& grid,int r,int c){
        int rr=grid.size();
        int cc=grid[0].size();

        grid[r][c]='0';//重新构建一个矩阵
        if(r-1>=0&&grid[r-1][c]=='1') dfs(grid,r-1,c);
        if(r+1<rr&&grid[r+1][c]=='1') dfs(grid,r+1,c);
        if(c-1>=0&&grid[r][c-1]=='1') dfs(grid,r,c-1);
        if(c+1<cc&&grid[r][c+1]=='1') dfs(grid,r,c+1);
    }
public:
    int numIslands(vector<vector<char>>& grid) {
        int rr=grid.size();
        if(!rr)return 0;
        int cc=grid[0].size();

        int num=0;
        for (int r=0;r<rr;r++) {
            for (int c=0;c<cc;c++){
                if (grid[r][c]=='1')  {
                    ++num;
                    dfs(grid,r,c);
                }
            }
        }
        return num;
    }
};