【leetcode with java】18 4Sum O(n^2)

我看了几个人气比较高的博客，他们这个算法都没做到O(n^2)，所以提前将我的解法贴出来分享，供大家参考（前面略过的题目近期都会补上的）。

【题目】：

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.
    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

    A solution set is:
    (-1,  0, 0, 1)
    (-2, -1, 1, 2)
    (-2,  0, 0, 2)

Tags: Array Hash Table Two Pointers

【解析】 我写代码的时候把思路写在了代码注释里了，请直接往下看：

【上码】

import java.util.ArrayList;
import java.util.Collections;
import java.util.Enumeration;
import java.util.Hashtable;
import java.util.List;
import java.util.Vector;

/*
 * 参考：未参考他人算法
 * O(n^2)
 * 思路：先将每对数塞进一个Hashtable（以这对数的和为key，以一个包含所有和为key的数对的集合为value）。
 * 然后遍历该Hashtable，对于每个和为key的数对集合，在该Hashtable中get和为target-key的数对集合，
 * 如果get到的为空，则continue，如果get到的不为空，则将这两个数对和并为一个有序4元组，并查看是否之前该
 * 4元组是否已经出现过（用另一个Hastable<String,AnyType>：reappear来快速判断，该reappear的
 * key是用有序4元组的4个整数拼接而来的）。 如果该4元组为出现过，则将其加入到结果集合里。
 * 
 */
public class Solution1 {
	public List<List<Integer>> fourSum(int[] num, int target) {
		List<List<Integer>> list = new ArrayList<List<Integer>>();
		if (num == null || num.length < 4)
			return list;
		int len = num.length;

		Hashtable<Integer,PairRecoder> prs = new Hashtable<Integer,PairRecoder>(len*len);
		for (int i = 0; i < len - 1; i++) {
			for (int j = i + 1; j < len; j++) {
				int sum = num[i]+num[j];
				PairRecoder pr = prs.get(num[i]+num[j]);
				if(pr == null)
					prs.put(sum, new PairRecoder(sum,i,j));
				else
					pr.pairs.add(new Pair(i,j));
			}
		}
		
		List<Four> fours = new ArrayList<Four>();
		Hashtable<String,Boolean> reappear = new  Hashtable<String,Boolean>(); 
		for(Enumeration<PairRecoder> e=prs.elements(); e.hasMoreElements(); ){
			PairRecoder pr1 = (PairRecoder)e.nextElement();
			PairRecoder pr2 = prs.get(target-pr1.v);
			if(pr2 == null)
				continue;
			for(int i=0;i<pr1.pairs.size();i++)
				for(int j=0;j<pr2.pairs.size();j++){
					Pair p1 = pr1.pairs.elementAt(i);
					Pair p2 = pr2.pairs.elementAt(j);
					if(p1.v1!=p2.v1 &&p1.v1!=p2.v2 && p1.v2!=p2.v1 && p1.v2!=p2.v2){
						Four of = new OrderedFour(num[p1.v1],num[p1.v2],num[p2.v1],num[p2.v2]);
						StringBuffer sb = new StringBuffer();
						sb.append(of.num.get(0));
						sb.append(of.num.get(1));
						sb.append(of.num.get(2));
						sb.append(of.num.get(3));
						if(!reappear.containsKey(sb.toString())){
							list.add(of.num);
							System.out.println("new："+sb);
							reappear.put(sb.toString(),true);
						}
					}
				}
		}
		
		return list;
	}

	class Pair {
		int v1;
		int v2;
		Pair(int v1, int v2) {
			this.v1 = v1;
			this.v2 = v2; 
		}
	}

	class PairRecoder{
		int v;
		Vector<Pair> pairs = new Vector<Pair>();
		PairRecoder(int v,int i,int j){
			this.v = v;
			this.pairs.add(new Pair(i,j));
		}
	}
	
	class Four{
		List<Integer> num;
		Four(int v1,int v2,int v3,int v4){
			this.num = new ArrayList<Integer>();
			num.add(v1);
			num.add(v2);
			num.add(v3);
			num.add(v4);
		}
	}
	class OrderedFour extends Four{
		OrderedFour(int v1,int v2,int v3,int v4) {
			super(v1, v2, v3, v4);
			Collections.sort(this.num);
		}
	}
}