200. Number of Islands ----- M

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return 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:

Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Example 2:

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

 

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] is '0' or '1'.

A:

 简单的dfs罢了。

根据fb面试的失败教训，建议每次发现一个‘1’， 先设‘0‘， 再dfs。

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