Number of Islands

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本文介绍了一个使用宽度优先搜索算法解决的岛屿计数问题。给定一个由'1'(陆地)和'0'(水域)组成的二维网格地图，通过连接相邻的'1'来形成岛屿，并计算岛屿总数。

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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

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

思路：宽度优先算法，碰到'1'没有被访问，则岛数++，同时该岛所有的'1'被标记为已访问。

public class Solution {
    LinkedList<int[]> q = new LinkedList<>();
    public int numIslands(char[][] grid) {
        if(grid==null){
            return 0;
        }
        int m=grid.length;
        if(m==0){
            return 0;
        }
        int n=grid[0].length;
        boolean[][] visited = new boolean[m][n];//标记(i,j)是否已被访问,这里仅标记'1',不标记'0'
        int cnt=0;//统计island的数目
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]=='1'&&visited[i][j]==false){//发现了新的island
                    bfs(grid,visited,i,j);//标记从该处'1'可达的所有'1'(即遍历整个岛屿)
                    cnt++;
                }
            }
        }
        return cnt;
    }
    private void bfs(char[][] grid, boolean[][] visited,int x,int y){
        fill(grid,visited,x,y);//先标记(x,y),再标记(x,y)的上下左右
        while(!q.isEmpty()){
            int[] point=q.poll();
            int i=point[0];
            int j=point[1];
            fill(grid,visited,i+1,j);
            fill(grid,visited,i-1,j);
            fill(grid,visited,i,j+1);
            fill(grid,visited,i,j-1);
        }
        
    }
    private void fill(char[][] grid, boolean[][] visited,int i,int j){
        if(i<0||i>=grid.length||j<0||j>=grid[0].length||grid[i][j]!='1'||visited[i][j]==true){
            return;
        }
        visited[i][j]=true;
        q.add(new int[]{i,j});
    }
    
}