LeetCode刷题实战200：岛屿数量

程序员小猿

发布于 2021-03-06 22:35:10

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发布于 2021-03-06 22:35:10

文章被收录于专栏：程序IT圈

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今天和大家聊的问题叫做 岛屿数量，我们先来看题面：

https://leetcode-cn.com/problems/number-of-islands/

Given an m x n 2d grid map of '1's (land) and '0's (water), return 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.

题意

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

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

解题

我们遍历整个二维数组，当我们遇到为'1'的，我们将其相邻的所有‘1’重置为‘0’，即将这一块岛全部都变为海水然后再去遍历数组剩下的部分。遇到一个岛屿的第一标‘1’的cnt加1。

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

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原始发表：2021-03-03，如有侵权请联系 cloudcommunity@tencent.com 删除

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