684. Redundant Connection

Medium

In this problem, a tree is an undirected graph that is connected and has no cycles.

You are given a graph that started as a tree with n nodes labeled from 1 to n, with one additional edge added. The added edge has two different vertices chosen from 1 to n, and was not an edge that already existed. The graph is represented as an array edges of length n where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the graph.

Return an edge that can be removed so that the resulting graph is a tree of n nodes. If there are multiple answers, return the answer that occurs last in the input.

Example 1:

Input: edges = [[1,2],[1,3],[2,3]]
Output:
 [2,3]

Example 2:

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

Constraints:

• n == edges.length

• 3 <= n <= 1000

• edges[i].length == 2

• 1 <= ai < bi <= edges.length

• ai != bi

• There are no repeated edges.

• The given graph is connected.

解題

看了好幾篇別人的解答參考，最好理解的是這篇

Runtime: 3 ms, faster than 93.75%

Memory Usage: 3.2 MB, less than 31.25%

func findRedundantConnection(edges [][]int) []int {
    parent := make([]int, 0) // 紀錄源頭

    for i := 0; i <= len(edges); i++ {
        parent = append(parent, i)
    }

    for _, edge := range edges {
        parent1 := find(edge[0], parent)
        parent2 := find(edge[1], parent)
        if parent1 == parent2 { // 已經有同一個源頭 = 早就相連
            return edge
        } 
        parent[parent2] = parent1 // 有相連的邊了，將不同源頭改為同一個
    }

    return []int{}
}

func find(n int, parent []int) int { // 找到源頭
    if parent[n] == n {
        return n
    } else {
        return find(parent[n], parent)
    }

    return 0
}