給你一個(gè)由 '1'（陸地）和 '0'（水）組成的的二維網(wǎng)格，請(qǐng)你計(jì)算網(wǎng)格中島嶼的數(shù)量。

島嶼總是被水包圍，并且每座島嶼只能由水平方向和/或豎直方向上相鄰的陸地連接形成。

此外，你可以假設(shè)該網(wǎng)格的四條邊均被水包圍。

示例 1：

輸入：grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
輸出：1

示例 2：

輸入：grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
輸出：3

提示：

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

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/number-of-islands
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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';//重新構(gòu)建一個(gè)矩陣
        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;
    }
};