README_EN.md

323. Number of Connected Components in an Undirected Graph

中文文档

Description

You have a graph of n nodes. You are given an integer n and an array edges where edges[i] = [ai, bi] indicates that there is an edge between ai and bi in the graph.

Return the number of connected components in the graph.

 

Example 1:

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

Example 2:

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

 

Constraints:

• 1 <= n <= 2000
• 1 <= edges.length <= 5000
• edges[i].length == 2
• 0 <= ai <= bi < n
• ai != bi
• There are no repeated edges.

Solutions

Python3

class Solution:
    def countComponents(self, n: int, edges: List[List[int]]) -> int:
        p = list(range(n))

        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        for a, b in edges:
            p[find(a)] = find(b)
        return sum(i == find(i) for i in range(n))

Java

class Solution {
    private int[] p;

    public int countComponents(int n, int[][] edges) {
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        for (int[] e : edges) {
            p[find(e[0])] = find(e[1]);
        }

        int cnt = 0;
        for (int i = 0; i < n; ++i) {
            if (i == find(i)) {
                ++cnt;
            }
        }
        return cnt;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

C++

class Solution {
public:
    vector<int> p;

    int countComponents(int n, vector<vector<int>>& edges) {
        p.resize(n);
        for (int i = 0; i < n; ++i) p[i] = i;
        for (auto e : edges) p[find(e[0])] = find(e[1]);
        int cnt = 0;
        for (int i = 0; i < n; ++i)
        {
            if (i == find(i)) ++cnt;
        }
        return cnt;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
};

Go

var p []int

func countComponents(n int, edges [][]int) int {
	p = make([]int, n)
	for i := 1; i < n; i++ {
		p[i] = i
	}
	for _, e := range edges {
		p[find(e[0])] = find(e[1])
	}
	cnt := 0
	for i := 0; i < n; i++ {
		if i == find(i) {
			cnt++
		}
	}
	return cnt
}

func find(x int) int {
	if p[x] != x {
		p[x] = find(p[x])
	}
	return p[x]
}

...