leetcode练习 Redundant Connection

这次是检查冗余的,题目主要练习无向图中是否有环的判断。
题目如下:
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.

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.

代码如下:

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        unordered_map<int, unordered_set<int>> m;
        for (auto x : edges) {
            if (ifCycle(x[0], x[1], m, -1)) return x;
            m[x[0]].insert(x[1]);
            m[x[1]].insert(x[0]);
        }
        return {};
    }

    bool ifCycle(int cur, int des, unordered_map<int, unordered_set<int>>&m, int pre) {
        if (m[cur].count(des)) return true;
        for (int num : m[cur]) {
            if (num == pre) continue;
            if (ifCycle(num, des, m, cur)) return true;
        }
        return false;
    }

};
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