[LeetCode] 200. Number of Islands Java

题目：

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

题意及分析：给出一个'1'和‘0’组成的二位矩阵，其中1代表陆地，0代表海洋，四个边界外都是海洋，要求计算有几块岛屿（岛屿定义为这片陆地四周都为海洋）。简单来讲就是要求几块独立的'1'组成的区域。对每个为‘1’的点进行宽度遍历（上下左右四个方向），遍历过的点置为‘2’，防止重复遍历。用quene记录该次遍历需要遍历的点。

代码：

public class Solution {
    public int numIslands(char[][] grid) {
        int row = grid.length;
        if(row==0) return 0;
        int col = grid[0].length;
        int count =  0;
        Queue<Integer[]> queue = new LinkedList<>();        ///记录已经被遍历的区域的点索引
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(grid[i][j]=='1'){      //找到某块为1的区域,遍历过后改为'2'
                    count++;　　　　//每次遇到新的区域便加1
                    Integer[] indexs={i,j};
                    queue.offer(indexs);
                    while(!queue.isEmpty()){
                        Integer[] temp = queue.poll();
                        int m=temp[0],n=temp[1];
                        if(grid[m][n]=='1'){
                            grid[m][n]='2';
                            if(m-1>=0&&grid[m-1][n]=='1'){
                                Integer[] index = {m-1,n};
                                queue.offer(index);
                            }
                            if(m+1<row&&grid[m+1][n]=='1'){
                                Integer[] index = {m+1,n};
                                queue.offer(index);
                            }
                            if(n-1>=0&&grid[m][n-1]=='1'){
                                Integer[] index = {m,n-1};
                                queue.offer(index);
                            }
                            if(n+1<col&&grid[m][n+1]=='1'){
                                Integer[] index = {m,n+1};
                                queue.offer(index);
                            }
                        }
                    }
                }
            }
        }
        return count;
    }
}