算法分析与设计丨第九周丨LeetCode（13）——Redundant Connection（Medium）

并查集算法

题目链接：https://leetcode.com/problems/redundant-connection/description/

题目描述：

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.

题目解析：这题也容易看出用并查集算法来做，但是速度并不算快，或许有更好的算法。做法其实和经典的并查集算法一致。

class Solution {
public:
int find(int x,vector<int>& father)
{
while(x!=father[x])
x = father[x];
return x;
}

vector<int> findRedundantConnection(vector<vector<int>>& edges) {
int size = edges.size();
vector<int> father(size+1);
vector<int> rank(size+1);

for(int i = 1;i<=size;++i)
{
father[i] = i;
rank[i] = 0;
}

vector<int> result;

for(int i = 0;i<size;++i)
{

int father_1 = find(edges[i][0],father);
int father_2 = find(edges[i][1],father);

if(father_1 == father_2)
{
result.push_back(edges[i][0]);
result.push_back(edges[i][1]);
continue;
}
else
{
if(rank[father_1] > rank[father_2])
{
father[father_2] = father_1;
}
else
{
if(rank[father_1] == rank[father_2])
rank[father_2]++;
father[father_1] = father_2;

}

}

}
return result;

}
};