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。
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