323. 无向图中连通分量的数目 🔒

English Version

题目描述

你有一个包含 n 个节点的图。给定一个整数 n 和一个数组 edges ，其中 edges[i] = [ai, bi] 表示图中 ai 和 bi 之间有一条边。

返回 图中已连接分量的数目 。

 

示例 1:

输入: n = 5, edges = [[0, 1], [1, 2], [3, 4]]
输出: 2

示例 2:

输入: n = 5, edges = [[0,1], [1,2], [2,3], [3,4]]
输出:  1

 

提示：

• 1 <= n <= 2000
• 1 <= edges.length <= 5000
• edges[i].length == 2
• 0 <= ai <= bi < n
• ai != bi
• edges 中不会出现重复的边

解法

方法一

class Solution:
    def countComponents(self, n: int, edges: List[List[int]]) -> int:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

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

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) {
            int a = e[0], b = e[1];
            p[find(a)] = find(b);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (i == find(i)) {
                ++ans;
            }
        }
        return ans;
    }

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

class Solution {
public:
    int countComponents(int n, vector<vector<int>>& edges) {
        vector<int> p(n);
        iota(p.begin(), p.end(), 0);
        for (int i = 0; i < n; ++i) p[i] = i;
        function<int(int)> find = [&](int x) -> int {
            if (p[x] != x) p[x] = find(p[x]);
            return p[x];
        };
        for (auto& e : edges) {
            int a = e[0], b = e[1];
            p[find(a)] = find(b);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) ans += i == find(i);
        return ans;
    }
};

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

/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {number}
 */
var countComponents = function (n, edges) {
    let p = new Array(n);
    for (let i = 0; i < n; ++i) {
        p[i] = i;
    }
    function find(x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
    for (const [a, b] of edges) {
        p[find(a)] = find(b);
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        if (i == find(i)) {
            ++ans;
        }
    }
    return ans;
};