题目地址：https://leetcode.com/problems/redundant-connection/description/
题目描述
Inthis problem, a tree is an undirected graph that is connected and has no cycles.
Thegiven 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.
Theresulting 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:
1、 Thesizeoftheinput2D-arraywillbebetween3and1000.；
2、 Everyintegerrepresentedinthe2D-arraywillbebetween1andN,whereNisthesizeoftheinputarray.；
题目大意
给出了一个有环无向图的各个边，找出能去除的边，使得这个图不含环（即为树）。
注意，这个图中的各个节点是1～N,总共有N条边，即只多了一条边。
解题方法
并查集
上一次看这个题的时候，我知道使用并查集去做，但是并没有做出来。这次再次心平气和的看的时候，已经能一遍写出来了。
关于并查集，这个知识点有点大。简而言之，告诉你一条边，去集合里查找这条边的两个节点分别属于哪个树。根据是否属于同一个树，做后续的判断。我之前的一篇文章讲述了并查集的一种应用：【九度OJ】题目1012：畅通工程 解题报告open in new window。更多的资料，可以看《计算机考研——机试指南》。
下面的代码实现了并查集查找根节点的代码，并且做了路径压缩，防止树太高导致查找根节点缓慢。
具体到这个题，虽然说是返回最后一个边，但我们知道只需要去除一条边就够了，之前的边不会构成环，直至多余的那条边出现。
另外要注意，当一条边的左右节点的根节点不同时，要把他们设置相同，这样等下次判断某条边的左右节点相同的情况时，说明是多余的那条边了。
python代码如下：
class Solution:
def findRedundantConnection(self, edges):
"""
:type edges: List[List[int]]
:rtype: List[int]
"""
tree = [-1] * (len(edges) + 1)
for edge in edges:
a = self.findRoot(edge[0], tree)
b = self.findRoot(edge[1], tree)
if a != b:
tree[a] = b
else:
return edge
def findRoot(self, x, tree):
if tree[x] == -1: return x
else:
root = self.findRoot(tree[x], tree)
tree[x] = root
return root
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C++代码如下：
class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
const int N = 1001;
m_ = vector<int>(N, 0);
for (int i = 0; i < N; ++i) {
m_[i] = i;
}
for (auto edge : edges) {
if (!u(edge[0], edge[1]))
return edge;
}
return {0, 0};
}
private:
vector<int> m_;
int f(int a) {
if (m_[a] != a)
m_[a] = f(m_[a]);
return m_[a];
}
bool u(int a, int b) {
int fa = f(a);
int fb = f(b);
if (fa == fb)
return false;
m_[fa] = b;
return true;
}
};
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