Number of Islands

标签：number of islands   leetcode   

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

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

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});
    }
    
}

Number of Islands

标签：number of islands   leetcode   

原文地址：http://blog.csdn.net/u010786672/article/details/44963827