Number of Islands

最新推荐文章于 2025-05-07 17:27:01 发布

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最新推荐文章于 2025-05-07 17:27:01 发布

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原文链接：http://www.cnblogs.com/ordili/p/4928517.html

本文介绍了一种算法，用于计算给定二维网格中被水包围的陆地数量。通过深度优先搜索（DFS）遍历网格，算法能够识别并计数相连的陆地，最终返回岛屿总数。

1. Title

Number of Islands

2. Http address

https://leetcode.com/problems/palindrome-linked-list/

3. The question

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

4. My code(AC)

 1     // Accepted
 2        public int numIslands(char[][] grid) {
 3              if ( grid == null || grid.length == 0 )
 4                  return 0;
 5             boolean visited[][] = new boolean[grid.length][grid[0].length];
 6             int count = 0;
 7             for(int i = 0 ; i < grid.length; i++ )
 8             {
 9                 for(int j = 0 ; j <grid[0].length; j++ )
10                 {
11                     if( grid[i][j] == '1' && visited[i][j] == false )
12                     {
13                         DFS(grid,i,j,visited);
14                         count++;
15                     }
16                 }
17             }
18 
19             return count;
20     
21         }
22         public  void DFS(char [][]grid, int i ,int j, boolean [][]visited){
23 
24                 if( visited[i][j] != false)
25                         return;
26                 int x = 0, y =0;
27                 visited[i][j] = true;
28                 //up
29                 x = i-1;
30                 y = j;
31                 if( x >= 0  && grid[x][y] == '1' && visited[x][y] == false ){
32                     DFS(grid, x, y ,visited);
33                 }
34                 
35                 //down
36                 x = i + 1;
37                 y = j;
38                 if( x < grid.length  && grid[x][y] == '1' && visited[x][y] == false ){
39                     DFS(grid, x, y ,visited);
40                 }
41 
42                 //left
43                 x = i;
44                 y = j-1;
45                 if( y >= 0 && grid[x][y] == '1' && visited[x][y] == false ){
46                     DFS(grid, x, y ,visited);
47                 }
48                 
49 
50                 //right
51                 x = i;
52                 y = j+1;
53                 if( y < grid[0].length  && grid[x][y] == '1' && visited[x][y] == false ){
54                     DFS(grid, x, y ,visited);
55                 }
56 
57         }