Number of Islands -- leetcode

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:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011

Answer: 3

基本思路：

1.遍历二维数组每个元素

2. 如果发现其为 ‘1‘， 则发现了一块岛屿，岛屿数量+1， 并从此点进行深度优先遍历，将该岛屿所有元素置上访问标记。

   递归遍历，即是向上、下、左、右，都进行访问。直到边界或者遇到水（‘0‘）、已访问（‘2）‘为止。

3. 最后将访问标记恢复为‘1‘。此步为可选，不恢复，leetcode也不会报错。

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int ans = 0;
        if (grid.empty() || grid[0].empty())
            return ans;
        
        const int m = grid.size();
        const int n = grid[0].size();
        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                if (grid[i][j] == '1') {
                    ++ans;
                    dfs(grid, i, j);
                }
            }
        }
        
        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++)
                if (grid[i][j] == '2')
                    grid[i][j] = '1';
        }
        return ans;
    }
    
    void dfs(vector<vector<char> >&grid, int i, int j) {
        if (j < 0 || i < 0 || 
            i == grid.size() || 
            j == grid[0].size() ||
            grid[i][j] != '1')
            return;
        
        grid[i][j] = '2';
        dfs(grid, i+1, j);
        dfs(grid, i-1, j);
        dfs(grid, i, j+1);
        dfs(grid, i, j-1);
    }
};

Number of Islands -- leetcode

原文：http://blog.csdn.net/elton_xiao/article/details/46793361