岛屿数量
作者:互联网
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/number-of-islands
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。
点击查看代码
class Solution {
public int numIslands(char[][] grid) {
//判断网格是否为空或者是否长度为 0 ,直接返回 0
if(grid == null || grid.length == 0){
return 0;
}
//定义网格边界
int nx = grid.length;
int ny = grid[0].length;
//创建记录岛屿数量的变量
int num = 0;
//利用坐标值遍历每一个网格
for(int x = 0;x < nx;x++){
for(int y = 0;y < ny;y++){
//判断当前网格值是否是 1
if(grid[x][y] == '1'){
//是 则岛屿数加一
num ++;
//调用递归方法,传入当前的网格数据
recursionIslands(grid,x,y);
}
}
}
//返回岛屿个数
return num;
}
//定义递归方法
public void recursionIslands(char[][] grid,int x,int y){
//定义边界
int nx = grid.length;
int ny = grid[0].length;
//判断如果超过边界或者网格值为 0 直接返回
if(x < 0 || x >= nx || y < 0 || y >= ny || grid[x][y] == '0'){
return;
}
//将当前网格值赋 0 ,避免后面重复递归
grid[x][y] = '0';
//递归当前网格上下左右的网格
recursionIslands(grid,x + 1,y);
recursionIslands(grid,x,y + 1);
recursionIslands(grid,x - 1,y);
recursionIslands(grid,x,y - 1);
}
}
示例 1:
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
示例 2:
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
标签:int,岛屿,网格,recursionIslands,length,grid,数量 来源: https://www.cnblogs.com/xy7112/p/16456559.html