<LeetCode OJ> 200. Number of Islands

Total Accepted: 48411 Total
Submissions: 171609 Difficulty: Medium

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:

11110110101100000000

Answer: 1

Example 2:

11000110000010000011

Answer: 3

Credits:

Special thanks to @mithmatt for adding this problem and creating all test cases.

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分析：

这是网上流传最广的深度优先解法，确实漂亮，通俗易理解。

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int result=0;
        if(grid.empty() || grid[0].empty())
            return result;
        int rows=grid.size();
        int cols=grid[0].size();
        for(int i=0;i<rows;i++)
        {
            for(int j=0;j<cols;j++)
            {
                if(grid[i][j]=='1')//如果是岛屿
                {
                    dfs(grid,i,j,rows,cols);//深搜，将该位置(i,j)四周的1都置0
                    result++;
                }
            }
        }
        return result;
    }
    void dfs(vector<vector<char>>& grid,int i,int j,int rows,int cols)
    {
        grid[i][j]='0';
        if(i > 0 && grid[i-1][j] == '1')//上边置0
            dfs(grid,i-1,j,rows,cols);
        if(i < rows-1 && grid[i+1][j] == '1')//下边同理
            dfs(grid,i+1,j,rows,cols);
        if(j > 0 && grid[i][j-1] == '1')//左边
            dfs(grid,i,j-1,rows,cols);
        if(j < cols-1 && grid[i][j+1] == '1')//右边
            dfs(grid,i,j+1,rows,cols);
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/51636977

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895