LeetCode 面试经典150题 [81/150 岛屿数量]


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GuoYulong 2024-07-31 132

题目描述

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 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

个人C++解答


递归搜索,果然一使用递归就有点不会了,说是深度优先搜索,但是没感觉到像

GYL
class Solution {
public:
    void dfs(vector<vector<char>>& grid, int i, int j) {
        int i_num = grid.size();
        int j_num = grid[0].size();

        grid[i][j] = '0';

        if (i - 1 >= 0 && grid[i - 1][j] == '1')
            dfs(grid, i - 1, j);
        if (i + 1 < i_num && grid[i + 1][j] == '1')
            dfs(grid, i + 1, j);
        if (j - 1 >= 0 && grid[i][j - 1] == '1')
            dfs(grid, i, j - 1);
        if (j + 1 < j_num && grid[i][j + 1] == '1')
            dfs(grid, i, j + 1);
    }
    int numIslands(vector<vector<char>>& grid) {
        int i_num = grid.size();
        if (!i_num)
            return 0;
        int j_num = grid[0].size();

        int ans = 0;
        for (int i = 0; i < i_num; i++) {
            for (int j = 0; j < j_num; j++) {
                if (grid[i][j] == '1') {
                    ++ans;
                    dfs(grid, i, j);
                }
            }
        }
        return ans;
    }
};

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