0684.Redundant Connection
Description
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
2
3
4
5
6
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
2
3
4
5
6
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.
Links
(en)https://leetcode.com/problems/redundant-connection/
(中文)https://leetcode-cn.com/problems/redundant-connection/
Solutions
Solution1
解法，使用 DisjoinSet 并查集实现的 Union-Find 来判断一个图是否连通。
/*
Complexity Analysis:
Time complexity: O(n).
Space complexity: O(n).
*/
public int[] findRedundantConnection(int[][] edges) {
int[] parent = new int[edges.length];
byte[] rank = new byte[edges.length];
for (int i = 0; i < edges.length; i++) {
parent[i] = i;
rank[i] = 0;
}
for (int[] edge : edges) {
if (connected(parent, edge[0]-1, edge[1]-1)) {
return edge;
} else {
union(parent, rank, edge[0]-1, edge[1]-1);
}
}
return null;
}
/**
* Returns the component identifier of the component containing site {@code p}.
*
* @param parent the root array of sites.
* @param p the integer representing one site.
* @return the component identifier of the component containing site {@code p}.
*/
private int find(int[] parent, int p) {
while (p != parent[p]) {
parent[p] = parent[parent[p]];
p = parent[p];
}
return p;
}
/**
* Returns true if two sites are in the same component.
* @param parent the root array of sites.
* @param p the integer representing one site.
* @param q the integer representing another site.
* @return {@code true} if two sites are in the same component;
* {@code false} otherwise.
*/
private boolean connected(int[] parent, int p, int q) {
return find(parent, p) == find(parent, q);
}
/**
* Merges the component containing site {@code p} with the component containing site {@code q}.
*
* @param parent the root array of sites.
* @param rank the rank of each site.
* @param p the integer representing one site.
* @param q the integer representing another site.
*/
private void union(int[] parent, byte[] rank, int p, int q) {
int rootP = find(parent, p);
int rootQ = find(parent, q);
if (rootP == rootQ) {
return;
}
// Always make the lower height tree to the higher tree.
if (rank[rootP] > rank[rootQ]) {
parent[rootQ] = rootP;
} else if (rank[rootP] < rank[rootQ]) {
parent[rootP] = rootQ;
}else {
parent[rootQ] = rootP;
rank[rootP]++;
}
}
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Submission status
提交的代码，运行时间 1 ms, 击败 78.35%。