leetcode 684. Redundant Connection (union-find演算法實現)
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.
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.
加權的union-find演算法,記錄每個根節點dep,這樣演算法find步驟就不會從對數複雜度?退化為線性複雜度。
時間複雜度分析見 https://leetcode.com/problems/redundant-connection/solution/
union-find的思想和dfs時的染色差不多,邊遍歷,邊標記。
class Solution {
public int[] findRedundantConnection(int[][] edges) {
UF uf=new UF(edges.length+1);
for(int[] e:edges){
if(uf.union(e[0],e[1])){
return e;
}
}
throw null;
}
class UF{
int[] id;//節點i聯通id[i],一直連線到根節點,為聯通的標識
int[] dep;//根節點深度
UF(int N){
id=new int[N];
dep=new int[N];
for(int i=0;i<N;i++) id[i]=i;
}
public int find(int n){
while(n!=id[n]) n=id[n];
return n;
}
public boolean union(int p,int q){
int pRoot=find(p);
int qRoot=find(q);
if(pRoot==qRoot) return true;
if(dep[pRoot]<dep[qRoot]){
id[pRoot]=qRoot;
}else if(dep[pRoot]>dep[qRoot]){
id[qRoot]=pRoot;
}else{
id[pRoot]=qRoot;
dep[qRoot]++;//深度+1
}
return false;
}
}
}