num of island 数岛屿 系列

1254. Number of Closed Islands Medium

Given a 2D grid consists of 0s (land) and 1s (water).  An island is a maximal 4-directionally connected group of 0s and a closed island is an island totally (all left, top, right, bottom) surrounded by 1s.

Return the number of closed islands.

Example 1:

Input: grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
Output: 2
Explanation: 
Islands in gray are closed because they are completely surrounded by water (group of 1s).

Example 2:

Input: grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
Output: 1

Example 3:

Input: grid = [[1,1,1,1,1,1,1],
               [1,0,0,0,0,0,1],
               [1,0,1,1,1,0,1],
               [1,0,1,0,1,0,1],
               [1,0,1,1,1,0,1],
               [1,0,0,0,0,0,1],
               [1,1,1,1,1,1,1]]
Output: 2

Constraints:

• 1 <= grid.length, grid[0].length <= 100
• 0 <= grid[i][j] <=1

解法：其中0的部分为island，但是接触到四个边界的island不算，只要按照这个规则进行dfs即可

class Solution {
    public int closedIsland(int[][] grid) {
        int count = 0;
        for( int i=0;i<grid.length;i++){
            for(int j=0;j<grid[0].length;j++){
                if(grid[i][j]==0 && dfs(grid,i,j)) {
                    count++;
                }
            }
        }
        return count;
    }
    private boolean dfs(int[][] grid,int i,int j){
        if(i<0||j<0||i>=grid.length||j>=grid[0].length) return false;
        if(grid[i][j]!=0) return true;
        grid[i][j]=1;
        return dfs( grid,i-1,j) & dfs(grid,i+1,j) & dfs(grid,i,j-1) & dfs(grid,i,j+1);
    }
}

 

标签：1s,int,island,岛屿,dfs,num,grid,Input
来源： https://www.cnblogs.com/cynrjy/p/15587068.html