[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 this combination in array S.

Attention:

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

2. Finally, the result is not tender, including 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, 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 arraylist of the last element in ascending order and does not appear repeated. Because each time it starts. The elements that are traversed are the smallest elements 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 repeated 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 repeated 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 heuristic traversal element in.                is the smallest of the 4 numbers, the Ans.add (target);                Ans.add (Num[l]); Ans.add (nuM[r]); Ret.add (ANS);                Put in the result set while (L < r && Num[l] = = num[l+1]) l++;//avoid repeated results, and it is directly skipped after it is equal to it.                while (L < r && Num[r] = = Num[r-1]) r--;////avoids repeated results, and it is directly skipped after it is equal to it.                l++;            r--;            } else if (Num[l] + num[r] + target < newtarget) l++;        else r--; }    }}

[Leetcode] [Java] 4Sum