Leetcode200岛屿数量(深搜）

问题描述

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
此外，你可以假设该网格的四条边均被水包围。

示例

输入：

grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]

输出：1
输入：

grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]

输出：3
 ##代码

class Solution {
  public int numIslands(char[][] grid) {
		 if(grid==null||grid.length==0||grid[0].length==0)
			 return 0;
		 	int num=0;
		 	for(int i=0;i<grid.length;i++) {
		 		for(int j=0;j<grid[0].length;j++) {
		 			if(grid[i][j]=='1') {
		 				num++;
		 				grid[i][j]='0';
		 				def(i,j,grid);
		 			}
		 		}
		 	}
		 	return num;
	    }
	void def(int i,int j,char[][]board) {
			if(i-1>=0&&board[i-1][j]=='1')
					{
					board[i-1][j]='0';
					def(i-1,j,board);
					}
			if(i+1<board.length&&board[i+1][j]=='1')
				{
				board[i+1][j]='0';
				def(i+1,j,board);
				}
			if(j-1>=0&&board[i][j-1]=='1')
				{
				board[i][j-1]='0';
				def(i,j-1,board);
				}
			if(j+1<board[0].length&&board[i][j+1]=='1')
				{
				board[i][j+1]='0';
				def(i,j+1,board);
				}
	}
}