[Leetcode] [Java] 4Sum

Title:

Given an array S of n integers, is 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 is 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)

Test Instructions:

Given an array of n integers, there are elements a,b,c and D in the array s, making a + b + C + D= target. Find all of these combinations in the array s.

Attention:

1. These four elements must be in ascending order(ie,a≤b≤C≤D)

2. The final result is not tender and contains repeated solutions.

    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)

Algorithm Analysis:

* Using the code of the title "3Sum" https://leetcode.com/problems/3sum/
* First the array is sorted in ascending order, and then the first element starts by traversing the 3Sum algorithm, the target value is (Target-nums[i]), and the iterated array is the array after the element
* This satisfies the final ArrayList in ascending order of the elements and does not duplicate, because each time it starts, the element that is traversed is the smallest element in the array

AC Code:

public class Solution {private static arraylist<arraylist<integer>> ret = new Arraylist<arraylist<in        Teger>> ();            Public arraylist<arraylist<integer>> foursum (int[] nums, int target) {int newtarget=0;        arraylist<arraylist<integer>> finalres = new arraylist<arraylist<integer>> ();    Arrays.sort (Nums); for (int i = 0; i < nums.length; i++) {arraylist<arraylist<integer>> temlist= new arraylist&lt ;    Arraylist<integer>> ();            if (i > 0 && nums[i] = = Nums[i-1]) continue;//Avoid duplicate results, and it is directly skipped after it is equal to it.            Newtarget=target-nums[i];            int inputnums[]=new int[nums.length-i-1]; for (int ii=0;ii<nums.length-i-1;ii++) inputnums[ii]=nums[ii+1+i];//each time the array following the element is selected as the input array of the 3SUM algorithm T                       Emlist=threesum (Inputnums,newtarget,nums[i]);            Finalres.addall (temlist);        Ret.clear (); }return FinalreS } public static arraylist<arraylist<integer>> Threesum (int[] num,int newtarget,int firstnum) {i                F (num = = NULL | | Num.length < 3) return RET;                Arrays.sort (num);        int len = num.length; for (int i = 0; i < len-2; i++) {if (i > 0 && num[i] = = Num[i-1]) continue;//Avoid duplicate results, behind and            It is equal to be skipped directly. Find (num, i+1, len-1, Num[i],newtarget, firstnum);    Look for two numbers with Num[i] and for 0} return ret; public static void Find (int[] num, int begin, int end, int target,int newtarget,int firstnum) {int L        = begin, R = end; while (L < R) {if (Num[l] + num[r] + target = = Newtarget) {arraylist<                integer> ans = new arraylist<integer> ();                Ans.add (firstnum);//different from the original 3Sum algorithm: Remember to add each of the revelation traversal elements, that is, the smallest of the 4 number of the Ans.add (target);                Ans.add (Num[l]); Ans.add (nuM[r]); Ret.add (ANS);                Put into the result set while (L < r && Num[l] = = num[l+1]) l++;//avoid duplication of results, and it is directly skipped after it is equal to it.                while (L < r && Num[r] = = Num[r-1]) r--;////avoids duplicate results, and is skipped directly after it is equal to it.                l++;            r--;            } else if (Num[l] + num[r] + target < newtarget) l++;        else r--; }    }}

Copyright NOTICE: This article is the original article of Bo Master, reprint annotated source

[Leetcode] [Java] 4Sum