标签：des style blog io ar color sp for java
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:
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)
这道题和3sum很像，所以我直接在3sum外面套了一层循环，开始以为会超时来着，因为时间复杂度为O(N^2*logn)但还是没超
开始用四重循环，穷举直接超时了，优化了一下也超时，就用的上面的方法
1 import java.util.ArrayList; 2 import java.util.Arrays; 3 import java.util.List; 4 5 6 7 8 9 //class ListNode { 10 // public int val; 11 // public ListNode next; 12 // ListNode(int x) { 13 // val = x; 14 // next = null; 15 // } 16 // } 17 18 public class Solution { 19 public List<List<Integer>> fourSum(int[] num, int target) { 20 List<List<Integer>> result = new ArrayList<List<Integer>>(); 21 Arrays.sort(num); //将数组升序排序 22 23 for(int q = 0; q < num.length;q++){ 24 if(q > 0 && num[q] == num[q - 1]) 25 continue; 26 for(int i = q + 1; i < num.length; i++){ 27 if(i > q + 1 && num[i] == num[i - 1]) 28 continue; //去重 29 int j = i + 1; 30 int k = num.length - 1; 31 while(j < k){ //从i + 1 ~ n - 1中找出两个数等于-num[i] 32 if(j > i + 1 && num[j] == num[j - 1]) 33 { 34 j++; 35 continue; 36 } 37 if(k < num.length - 1 && num[k] == num[k + 1]){ 38 k--; 39 continue; 40 } //去重 41 int temp = num[i] + num[j] + num[k] + num[q]; 42 if(temp > target){ //结果比0大k前移 43 k--; 44 continue; 45 } 46 else if(temp < target){ 47 j++; 48 continue; 49 }else{ //找到一个解 50 List<Integer> element = new ArrayList<Integer>(); 51 element.add(num[q]); 52 element.add(num[i]); 53 element.add(num[j]); 54 element.add(num[k]); 55 result.add(element); 56 j++; 57 //break; //退出循环 58 } 59 60 } 61 } 62 } 63 64 65 66 return result; 67 } 68 }
标签：des style blog io ar color sp for java
原文地址：http://www.cnblogs.com/luckygxf/p/4148770.html
