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200. Number of Islands

时间:2017-02-02 11:15:14      阅读:192      评论:0      收藏:0      [点我收藏+]

200. Number of Islands

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  • Difficulty: Medium
  • Contributors: Admin

 

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

DFS:

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

 

 

 

200. Number of Islands

原文:http://www.cnblogs.com/93scarlett/p/6360953.html

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