标签:xpl mono ati dir rect sam The 技术 class
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
class Solution { public int closedIsland(int[][] grid) { int m = grid.length; int n = grid[0].length; int res = 0; for(int i = 0; i < m; i++){ for(int j = 0; j < n; j++){ if(grid[i][j] == 0 && i == 0 || j == 0 || i == m - 1 || j == n - 1) help(i, j, grid); } } for(int i = 0; i < m; i++){ for(int j = 0; j < n; j++){ if(grid[i][j] == 0){ res++; help(i, j, grid); } } } return res; } public void help(int r, int c, int[][] grid){ if(r > -1 && r < grid.length && c > -1 && c < grid[0].length && grid[r][c] == 0) grid[r][c] = 1; else return; help(r + 1, c, grid); help(r, c + 1, grid); help(r - 1, c, grid); help(r, c - 1, grid); } }
DFS, 先把边界的0区域全置1,然后再遍历,遇到0就先加1,然后再把该0区域置1,最后得到答案。
1254. Number of Closed Islands
标签:xpl mono ati dir rect sam The 技术 class
原文地址:https://www.cnblogs.com/wentiliangkaihua/p/11870644.html