gpt4 book ai didi

323. Number of Connected Components in an Undirected Graph 无向图中连通分量的数目

转载 作者:大佬之路 更新时间:2024-01-31 14:20:27 24 4
gpt4 key购买 nike

题目地址:https://leetcode-cn.com/problems/number-of-connected-components-in-an-undirected-graph/

题目描述

Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to find the number of connected components in an undirected graph.

Example 1:

Input: n = 5 and edges = [[0, 1], [1, 2], [3, 4]]

     0          3
     |          |
     1 --- 2    4 

Output: 2

Example 2:

Input: n = 5 and edges = [[0, 1], [1, 2], [2, 3], [3, 4]]

     0           4
     |           |
     1 --- 2 --- 3

Output:  1

Note:

Youcan assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

题目大意

给定编号从 0 到 n-1 的 n 个节点和一个无向边列表(每条边都是一对节点),请编写一个函数来计算无向图中连通分量的数目。

解题方法

并查集

看到求联通分量的题,一般都可以用并查集。比如1101. The Earliest Moment When Everyone Become Friendsopen in new window

只要把并查集背下来,这个题目基本直接写上去就好了。

C++代码如下:

class Solution {
public:
    int countComponents(int n, vector<vector<int>>& edges) {
        map_ = vector<int>(n, 0);
        components = n;
        for (int i = 0; i < n; ++i) {
            map_[i] = i;
        }
        for (vector<int>& edge : edges) {
            uni(edge[0], edge[1]);
        }
        return components;
    }
    int find(int a) {
        if (a == map_[a])
            return a;
        return find(map_[a]);
    }
    void uni(int a, int b) {
        int pa = find(a);
        int pb = find(b);
        if (pa == pb)
            return;
        map_[pa] = pb;
        components --;
    }
private:
    vector<int> map_;
    int components;
};

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

2022

DDKK.COM 弟弟快看-教程,程序员编程资料站,版权归原作者所有

本文经作者:负雪明烛 授权发布,任何组织或个人未经作者授权不得转发

24 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com