Code Interview Note
  • 0. Introduction
  • 1. Basic
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  • 2. Array and Numbers
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  • 2.1 String
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    • Calculator
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  • 3. Linkedlist
    • Dummy Node
    • Double Pointers
  • 4. Stack and Queue
    • General
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  • 5. Binary Search
    • General
    • BS on result
    • Save the half which has result
    • Rotated Sorted Array
    • Split Array Largest Sum
  • 6. Binary Tree
    • General
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    • Lowest Common Ancestor
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    • General
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  • 9. DFS
    • Backtrack
    • General
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  • 10. HashTable
    • General
  • 11. Sort
    • General
  • 12. Recursion
    • General
  • 13. Dynamic Programming
    • Graph
    • General
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    • Double Sequence
    • Longest Common Subsequence
    • Rolling Array
    • House Robber
    • Backpack
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  • 14. BFS
    • General
    • Number of Islands
    • The Maze
  • 15. Graph
    • Shortest Path
    • Undirected Graph
    • Topology Sort
    • Word Ladder
    • Tarjan's Algo
  • 16. Divide & Conquer
    • General
  • 17. UnionFind
    • General
    • Grouping
  • 18. Trie
    • General
    • Word Square
  • 19. Company Summary
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17. UnionFind

PreviousGeneralNextGeneral

Last updated 5 years ago

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  • 算法介紹:

  • 合併集合

  • 判斷元素是否在同一個集合中

public class unionFind{
    // 紀錄上一層的node
    private int[] father = null;
    // 記錄有幾個組
    private int count;
    // 記錄每組有幾個元素
    private int[] size = null;
    // initiate data structure
    public unionFind(int n){
        count = n;
        father = new int[n];
        size = new int[n];
        for(int i=0; i<n; i++){
            father[i] = i;
            size[i] = 1;
        }
    }
    // path compression
    private find(int x){
        if(father[x] != x) father[x] = find(father[x]);
        return father[x];
    }

    public void union(int p, int q){
        int root_p = find(p);
        int root_q = find(q);
        if(root_p == root_q) return;
        if(size(root_p) > size(root_q)){
            father[root_q] = root_p;
            size[root_p] += size[root_q];

        } 
        else{
            father[root_p] = root_q; 
            size[root_q] += size[root_p];
        }
        count--; 
    }
https://blog.csdn.net/dm_vincent/article/details/7655764