标签：ace div ++ 访问 遍历 比较 connect solution may
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:
Input: 11110 11010 11000 00000 Output: 1
Example 2:
Input: 11000 11000 00100 00011 Output: 3
大神的代码，比较好。将遍历到的点全部变成0，这样就不用visit数组了来记录是否访问过了
public class Solution { private int n; private int m; public int numIslands(char[][] grid) { int count = 0; n = grid.length; if (n == 0) return 0; m = grid[0].length; for (int i = 0; i < n; i++){ for (int j = 0; j < m; j++) if (grid[i][j] == ‘1‘) { DFSMarking(grid, i, j); ++count; } } return count; } private void DFSMarking(char[][] grid, int i, int j) { if (i < 0 || j < 0 || i >= n || j >= m || grid[i][j] != ‘1‘) return; grid[i][j] = ‘0‘; DFSMarking(grid, i + 1, j); DFSMarking(grid, i - 1, j); DFSMarking(grid, i, j + 1); DFSMarking(grid, i, j - 1); } }
我的代码，略微麻烦
class Solution { int[][] visit; int ans=0; public int numIslands(char[][] grid) { if(grid.length==0) return 0; visit=new int[grid.length][grid[0].length]; for(int i=0;i<grid.length;i++){ for(int j=0;j<grid[0].length;j++){ if(grid[i][j]==‘1‘&&visit[i][j]!=1){ ans++; fight(i,j,grid); } } } return ans; } public void fight(int i,int j,char[][] grid){ visit[i][j]=1; if(grid[i][j]==‘0‘) return; else{ if(i+1<grid.length&&visit[i+1][j]!=1) fight(i+1,j,grid); if(i-1>=0&&visit[i-1][j]!=1) fight(i-1,j,grid); if(j+1<grid[0].length&&visit[i][j+1]!=1) fight(i,j+1,grid); if(j-1>=0&&visit[i][j-1]!=1) fight(i,j-1,grid); } } }
leetcode--200. Number of Islands
标签：ace div ++ 访问 遍历 比较 connect solution may
原文地址：https://www.cnblogs.com/albert67/p/10419748.html
