LN : leetcode 684 Redundant Connection
lc 684 Redundant Connection
684 Redundant Connection
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]
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.
Union Find Accepted
Union Find的關鍵思想是使各結點依次連結在一起,如果有從1到2的邊,則令uni[1] = 2
,如果有從2到3的邊,則令uni[2] = 3
,這樣如果此時新加一條從1到3的邊,那麽從1開始尋找,uni1為2,uni[2]為3,即1已經有了一條從1到3的邊,再加會導致形成環路,所以這就是我們要的答案。
class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
vector<int> uni(2001, -1);
for (auto edge : edges) {
int head = edge[0], tail = edge[1];
int x = find(uni, head), y = find(uni, tail);
if (x == y) return edge;
uni[x] = y;
}
return {};
}
int find(vector<int>& uni, int num) {
while(uni[num] != -1) {
num = uni[num];
}
return num;
}
};
LN : leetcode 684 Redundant Connection