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剑指 Offer II 118. 多余的边

题目描述

树可以看成是一个连通且 无环 的 无向 图。

给定往一棵 n 个节点 (节点值 1～n) 的树中添加一条边后的图。添加的边的两个顶点包含在 1 到 n 中间，且这条附加的边不属于树中已存在的边。图的信息记录于长度为 n 的二维数组 edges ，edges[i] = [ai, bi] 表示图中在 ai 和 bi 之间存在一条边。

请找出一条可以删去的边，删除后可使得剩余部分是一个有着 n 个节点的树。如果有多个答案，则返回数组 edges 中最后出现的边。

 

示例 1：

输入: edges = [[1,2],[1,3],[2,3]]
输出: [2,3]

示例 2：

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

 

提示:

  • n == edges.length
  • 3 <= n <= 1000
  • edges[i].length == 2
  • 1 <= ai < bi <= edges.length
  • ai != bi
  • edges 中无重复元素
  • 给定的图是连通的 

 

注意：本题与主站 684 题相同： https://leetcode.cn/problems/redundant-connection/

解法

方法一

Python3

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

        p = list(range(1010))
        for a, b in edges:
            if find(a) == find(b):
                return [a, b]
            p[find(a)] = find(b)
        return []

Java

class Solution {
    private int[] p;

    public int[] findRedundantConnection(int[][] edges) {
        p = new int[1010];
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
        }
        for (int[] e : edges) {
            int a = e[0], b = e[1];
            if (find(a) == find(b)) {
                return e;
            }
            p[find(a)] = find(b);
        }
        return null;
    }

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

C++

class Solution {
public:
    vector<int> p;

    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        p.resize(1010);
        for (int i = 0; i < p.size(); ++i) p[i] = i;
        for (auto& e : edges) {
            int a = e[0], b = e[1];
            if (find(a) == find(b)) return e;
            p[find(a)] = find(b);
        }
        return {};
    }

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

Go

func findRedundantConnection(edges [][]int) []int {
	p := make([]int, 1010)
	for i := range p {
		p[i] = i
	}
	var find func(x 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]
		if find(a) == find(b) {
			return []int{a, b}
		}
		p[find(a)] = find(b)
	}
	return []int{}
}

Swift

class Solution {
    private var parent: [Int] = []

    func findRedundantConnection(_ edges: [[Int]]) -> [Int] {
        parent = Array(0..<1010)

        for edge in edges {
            let a = edge[0]
            let b = edge[1]

            if find(a) == find(b) {
                return edge
            }
            parent[find(a)] = find(b)
        }
        return []
    }

    private func find(_ x: Int) -> Int {
        if parent[x] != x {
            parent[x] = find(parent[x])
        }
        return parent[x]
    }
}