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:
Update (2017-09-26):
We have overhauled the problem description + test cases and specified clearly the graph is an undirected graph. For the directed graph follow up please see Redundant Connection II). We apologize for any inconvenience caused.
A:
因为u<v, 很自然想到union 的思路。
可是, merge的时候,我们还需要 再map back 回来。 因为不能简单地用指针都指向同一个地址。 (会给后续的修改 造成missing ends)
为了简单快捷,我这里用了for loop
class Solution { public: vector<int> findRedundantConnection(vector<vector<int>>& edges) { int N = edges.size(); vector<int> V(N+1, INT_MAX); // node value as index, V[i] is their smallest parent for (auto edge : edges) { int u = edge[0]; int v = edge[1]; int smallest = min(u, min(V[u], V[v]) ); if(V[u] != INT_MAX && V[u] == V[v]) // if already same parent return vector<int>{u,v}; int preUsmallest = V[u]; int preVsmallest = V[v]; for(int i =1;i<=N;i++){ if(i==u || i==v || (V[i] != INT_MAX && (V[i] == preUsmallest || V[i] == preVsmallest) )){// if parent V[i] = smallest; } } } return vector<int>();// shall not hit } };
No comments:
Post a Comment