岛屿数量

2022/7/8 6:22:52

本文主要是介绍岛屿数量,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!

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



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