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Copy pathConnecting Graph III.java
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81 lines (69 loc) · 1.89 KB
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1520323016
tags: Union Find
还是UnionFind的变形, 这次是算有剩下多少个union. 其实非常简单, 维持一个全局变量count:
一开始count=n, 因为全是散装elements; 每次union, count--.
```
/*
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(), Returns the number of connected component in the graph
Example
5 // n = 5
query() return 5
connect(1, 2)
query() return 4
connect(2, 4)
query() return 3
connect(1, 4)
query() return 3
*/
/**
Thoughts:
Not able to run on LintCode.
Count the number of unions left
*/
public class ConnectingGraph {
// Placeholder for all the UninFind relationships
private int[] father = null;
private int count;
/**
Initialize one element to each individual union.
*/
public ConnectingGraph(int n) {
father = new int[n + 1];
count = n;
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) {
count--;
parent[a] = rootB; // doesn't mater which assigns to which one.
}
}
/** Return count of union left */
public boolean query() {
return count;
}
/*
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]);
}
}
```