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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)
类似3Sum方法，只是在外层多加一个for循环，但是对于去重的控制边界条件需要注意，这里的while中n的判断下线为j+1
1 package si.Sum; 2 3 import java.util.ArrayList; 4 import java.util.Arrays; 5 import java.util.List; 6 7 public class SiSum { 8 public static void main(String args[]){ 9 //int num[]={2,1,0,-1}; 10 int num[]={-1,0,-5,-2,-2,-4,0,1,-2}; 11 List<List<Integer>> list=fourSum(num, -9); 12 for(List<Integer> temp:list){ 13 for(int i:temp){ 14 System.out.print(i+"-->"); 15 } 16 System.out.println(); 17 } 18 } 19 public static List<List<Integer>> fourSum(int[] num, int target) { 20 Arrays.sort(num); 21 int len=num.length; 22 List<List<Integer>> list=new ArrayList<List<Integer>>(); 23 for(int i=0;i<=len-4;i++){ 24 for(int j=i+1;j<=len-3;j++){ 25 int m=j+1; 26 int n=len-1; 27 int diff=(target-num[i]-num[j]); 28 while(m<n){ 29 int twoSum=num[m]+num[n]; 30 if(twoSum>diff){ 31 n--; 32 }else if(twoSum<diff){ 33 m++; 34 }else{ 35 List<Integer> tempList=new ArrayList<Integer>(); 36 tempList.add(num[i]); 37 tempList.add(num[j]); 38 tempList.add(num[m]); 39 tempList.add(num[n]); 40 list.add(tempList); 41 n--; 42 m++; 43 while(m<len-1&&num[m]==num[m-1]) 44 { 45 m++; 46 } 47 while(n>=j+1&&n<len-1&&num[n]==num[n+1]){ 48 n--; 49 } 50 } 51 } 52 while(j<len-1&&num[j+1]==num[j]){ 53 j++; 54 } 55 } 56 while(i<len-1&&num[i+1]==num[i]){ 57 i++; 58 } 59 } 60 return list; 61 } 62 }
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原文地址：http://www.cnblogs.com/criseRabbit/p/4202244.html
