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Leetcode 200. Number of Islands

时间:2016-05-20 23:54:13      阅读:151      评论:0      收藏:0      [点我收藏+]

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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

 

技术分享
 1 class Solution {
 2 public:
 3     int numIslands(vector<vector<char>>& grid) {
 4         int many = 0;
 5         if(grid.empty() || grid[0].empty())
 6             return many;
 7         //void dfs(vector<vector<char>> &grid,int i, int j, int n, int m);
 8         int n = grid.size();
 9         int m = grid[0].size();
10         for(int i = 0; i < n; i++){
11             for(int j = 0; j < m; j++){
12                 if(grid[i][j] == 1){
13                     dfs(grid,i,j,n,m);
14                     many++;
15                 }
16             }
17         }
18         return many;
19     }
20     void dfs(vector<vector<char>>& grid,int i, int j, int n, int m)
21     {
22         grid[i][j] = 0;
23         if((i-1) >= 0 && grid[i-1][j] == 1)
24             dfs(grid, i-1, j, n, m);
25         if((i+1) < n && grid[i+1][j] == 1)
26             dfs(grid, i+1, j, n, m); 
27         if((j-1) >= 0 && grid[i][j-1] == 1)
28             dfs(grid, i, j-1, n, m);
29         if((j+1) < m && grid[i][j+1] == 1)
30             dfs(grid, i, j+1, n, m);
31     }
32 };
View Code

 

Leetcode 200. Number of Islands

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原文地址:http://www.cnblogs.com/qinduanyinghua/p/5513692.html

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