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1520320428
tags: Union Find
没有跑过这个程序, 是一个UnionFind的简单实现.
Document了每个环节的计算原理/思想.
```
/*
LintCode
Given n nodes in a graph labeled from 1 to n. There is no edges in the graph at beginning.
You need to support the following method:
1. connect(a, b), add an edge to connect node a and node b.
2. query(a, b), check if two nodes are connected
Example
5 // n = 5
query(1, 2) return false
connect(1, 2)
query(1, 3) return false
connect(2, 4)
query(1, 4) return true
*/
/*
Thoughts:
Not tested on lintcode.
Implementation of Union Find
*/
public class ConnectingGraph {
// Placeholder for all the UninFind relationships
private int[] father = null;
/**
Initialize one element to each individual union.
*/
public ConnectingGraph(int n) {
father = new int[n + 1];
for (int i = 1; i <= n; i++) {
father[i] = i;
}
}
/**
Union function.
Find the root father, if not the same, union them together.
*/
public void connect(int a, int b) {
int rootA = find(a);
int rootB = find(b);
if (rootA != rootB) {
parent[a] = rootB; // doesn't mater which assigns to which one.
}
}
/** Check if the two integer are in the same union */
public boolean query(int a, int b) {
return find(a) == find(b);
}
/*
Find function: find the root parent as the head for entire union.
If found parent as itself, return it.
Otherwise, recursively look for father and assign the result eventually.
*/
private int find(int x) {
if (father[x] == x) {
return x; // x is the root parent, return itself.
}
return father[x] = find(father[x]);
}
}
```