【leetcode with java】18 4Sum O(n^2)

来源：互联网发布：淘宝机油之家编辑：程序博客网时间：2024/05/28 06:07

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));elsepr.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);}}}

0 0