【LeetCode】684. Redundant Connection

问题描述

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.

思路

判断无向图是否有环有几种方法：

1.用并查集来判断

        若两个点的根节点相同（这两个点是父子关系），则形成环;
        若所有点中只有一个点的根节点是他本身，则根节点入度为0。

2.DFS搜索图，图中的边只可能是树边或反向边，一旦发现反向边，则表明存在环。

3.如果存在回路，则必存在一个子图，是一个环路。环路中所有顶点的度>=2。

       第一步：删除所有度<=1的顶点及相关的边，并将另外与这些边相关的其它顶点的度减一。
         第二步：将度数变为1的顶点排入队列，并从该队列中取出一个顶点重复步骤一。
         如果最后还有未删除顶点，则存在环，否则没有环。

在这道题中，我们采用第一种方法。从题目看，每个数字代表一个点 ，根据第一个方法，我们需要找到第一条起点和终点有一个相同根节点的边，这条边即为结果。

初始化时，所有的点将自己作为自己的根节点。然后开始遍历edges，更新遍历的点的根节点直到找到环为止。

代码

int findParent(vector<int>& parent, int k) {    if (parent[k] != k)         parent[k] = findParent(parent, parent[k]);    return parent[k];}vector<int> findRedundantConnection(vector<vector<int> >& edges) {vector<int> parent;    for (int i = 0; i < 2001; i++)  // 初始化    parent.push_back(i);    int point1, point2;   for (int j = 0; j < edges.size(); j++) {   point1 = findParent(parent, edges[j][0]);   point2 = findParent(parent, edges[j][1]);        if (point1 == point2)        return edges[j];        parent[point2] = point1;   }    return vector<int>(0, 0);}