https://leetcode.com/problems/redundant-connection/description/
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.
显然是个并查集的题。
直接初始化并查集，每碰到一条边合并集合即可。
class Solution {
public:
int fa[1005];
bool unionTest(int u, int v) {
while (fa[u] != u) u = fa[u];
while (fa[v] != v) v = fa[v];
if (u != v) {
fa[v] = u;
return false;
} else {
return true;
}
}
vector findRedundantConnection(vector>& edges) {
int n = edges.size();
int ansu, ansv;
ansu = ansv = 0;
for (int i = 1; i <= n; i++) {
fa[i] = i;
}
for (int i = 0; i < n; i++) {
int u = edges[i][0];
int v = edges[i][1];
if (unionTest(u, v)) {
ansu = u;
ansv = v;
}
}
vector ans;
ans.push_back(ansu);
ans.push_back(ansv);
return ans;
}
};