LeetCode #684 Redundant Connection

xiaoxiao2021-02-28  24

题目

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

The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, …, 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 resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] with u < v, that represents an undirected edge connecting nodes u and v.

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 given 2D-array. The answer edge [u, v] should be in the same format, with u < v.

Example 1:

Input: [[1,2], [1,3], [2,3]] Output: [2,3] Explanation: The given undirected graph will be like this: 1 / \ 2 - 3

Example 2:

Input: [[1,2], [2,3], [3,4], [1,4], [1,5]] Output: [1,4] Explanation: The given undirected graph will be like this: 5 - 1 - 2 | | 4 - 3

Note:

The size of the input 2D-array will be between 3 and 1000.Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

解题思路

这道题用并查集可以很容易地解决。 遍历每一个 edge ,如果第二个顶点所在的树的根结点与第一个结点的根结点不相同,说明在当前构建的树中还不存在路径把两个顶点连接起来,则把当前 edge 的第二个顶点所在的树合并到第一个顶点的树中,使这两个顶点具有相同的根结点,这样就表示找到了一条路径可以连接当前 edge 的两个顶点; 如果第二个顶点所在的树的根结点与第一个结点的根结点相同,说明在前面的构建树的过程中,已经找到路径连接这两个顶点了,因此当前的这个 edge 就是冗余的,可以删掉。把当前的 edge 作为结果返回即可。 (注意一旦找到了冗余的 edge ,则这个 edge 肯定就是 the answer that occurs last in the given 2D-array,用题目的例子细细推导一下就可以知道)

C++代码实现

class Solution { public: vector<int> findRedundantConnection(vector<vector<int>>& edges) { int N = edges.size(); unordered_map<int, int> parents; vector<int> rst; for (int i = 1; i <= N; ++i) { parents[i] = i; } for (int i = 0; i < N; ++i) { int firstP = findParent(edges[i][0], parents); int secondP = findParent(edges[i][1], parents); if (firstP == secondP) { rst.push_back(edges[i][0]); rst.push_back(edges[i][1]); break; } else { parents[secondP] = firstP; } } return rst; } int findParent(int son, unordered_map<int, int>& parents) { if (son == parents[son]) { return son; } parents[son] = findParent(parents[son], parents); return parents[son]; } };
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