原題鏈接在這里：https://leetcode.com/problems/redundant-connection/

題目：

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?undirectededge 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.

題解：

For each edge, union edge[0] and edge[1]. If edge[0] and edge[1] are already in the same union. Then current edge is redundant.

Time Complexity: O(nlogn). find takes O(logn). With path compression and union by weight, amatorize O(1).

Space: O(n).

AC Java:

1 class Solution { 2 int [] parent; 3 int [] size; 4 5 public int[] findRedundantConnection(int[][] edges) { 6 int n = edges.length; 7 parent = new int[n+1]; 8 size = new int[n+1]; 9 10 for(int i = 0; i<=n; i++){ 11 parent[i] = i; 12 size[i] = 1; 13 } 14 15 for(int [] edge: edges){ 16 if(find(edge[0]) == find(edge[1])){ 17 return edge; 18 } 19 20 union(edge[0], edge[1]); 21 } 22 23 return null; 24 } 25 26 private int find(int i){ 27 if(i != parent[i]){ 28 parent[i] = find(parent[i]); 29 } 30 31 return parent[i]; 32 } 33 34 private void union(int i, int j){ 35 int p = find(i); 36 int q = find(j); 37 38 if(size[p] > size[q]){ 39 parent[q] = p; 40 size[p] += size[q]; 41 }else{ 42 parent[p] = q; 43 size[q] += size[p]; 44 } 45 } 46 }

跟上Redundant Connection II.

轉載于:https://www.cnblogs.com/Dylan-Java-NYC/p/11236746.html

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