[LeetCode]Number of Islands

2015-04-08 19:59 庸男勿扰阅读(194) 评论(0) 编辑收藏举报

原题链接：https://leetcode.com/problems/number-of-islands/

题意描述：

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:

Answer: 1

Example 2:

Answer: 3

题解：

此题的大意就是找出一个图中的联通区域，一个联通子图便认为是一个岛屿。显然就是对每一个未访问过且是陆地的点进行上下左右搜索，这样即可找到一个联通子图。然后对全图都这样做一遍，便能找到所有的，即结果。代码如下：

 1 public class Solution {
 2     public static boolean flag = false;
 3 
 4     public int numIslands(char[][] grid) {
 5         boolean[][] visited = new boolean[grid.length][];
 6         for (int i = 0; i < grid.length; i++)
 7             visited[i] = new boolean[grid[i].length];
 8         for (int i = 0; i < grid.length; i++) {
 9             for (int j = 0; j < grid[0].length; j++) {
10                 visited[i][j] = false;
11             }
12         }
13         int res = 0;
14         for (int i = 0; i < grid.length; i++) {
15             for (int j = 0; j < grid[0].length; j++) {
16                 flag = false;
17                 process(i, j, grid, visited);
18                 if (flag)
19                     res++;
20             }
21         }
22         return res;
23     }
24 
25     public void process(int i, int j, char[][] grid, boolean[][] visited) {
26         if (grid[i][j] != '1' || visited[i][j])
27             return;
28         visited[i][j] = true;
29         flag = true;
30         if (j >= 1)
31             process(i, j - 1, grid, visited);
32         if (i >= 1)
33             process(i - 1, j, grid, visited);
34         if(i+1<grid.length)
35             process(i + 1, j, grid, visited);
36         if(j+1<grid[0].length)
37             process(i, j + 1, grid, visited);
38     }
39 }