leetcode--Number of Islands_java leetcode number of islands 非递归-优快云博客

本文介绍了一种计算二维网格中岛屿数量的方法。通过遍历网格，使用广度优先搜索（BFS）来标记并计数岛屿的数量。算法适用于由'1'（陆地）和'0'（水域）组成的地图。

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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:

Answer: 1

Example 2:

Answer: 3

[java]view plaincopy 
   
 public class Solution {  
     public int numIslands(char[][] grid) {  
         int res = 0;  
         int rows = grid.length;  
         if(rows==0) return res;  
         int cols = grid[0].length;  
         if(cols==0) return res;  
         int[] xdir = new int[]{1,-1,0,0};  
         int[] ydir = new int[]{0,0,1,-1};  
         ArrayList<Integer> queue_x = new ArrayList<Integer>();  
         ArrayList<Integer> queue_y = new ArrayList<Integer>();  
           
         for(int i=0;i<rows;i++){  
             for(int j=0;j<cols;j++){  
                 if(grid[i][j]=='1'){  
                     res++;  
                     queue_x.add(i);queue_y.add(j);  
                     int low = 0;  
                     int high = 1;  
                     while(low<high){  
                         for(int k=0;k<4;k++){  
                             int x = queue_x.get(low)+xdir[k];  
                             int y = queue_y.get(low)+ydir[k];  
                             if(x>=0&&x<rows&&y>=0&&y<cols){  
                                 if(grid[x][y]=='1'){  
                                     grid[x][y]='0';  
                                     queue_x.add(x);  
                                     queue_y.add(y);  
                                     high++;  
                                 }  
                             }  
                         }  
                         low++;  
                     }  
                     queue_x.clear();queue_y.clear();  
                 }  
             }  
         }  
         return res;  
     }  
 }  