# "Leetcode-Interview algorithm classic-java Implementation" "018-4sum (four numbers and)"

Source: Internet
Author: User

"018-4sum (four-digit and)" "leetcode-Interview algorithm classic-java Implementation" "All topics Directory Index" Original Question

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)``

Main Topic

Given an array of integers, find the unique solution for A + B + c + d = target.

Thinking of solving problems

First determine the two numbers of a and d, for a and D two numbers, cannot be reused at the same time. Then determine B and C, and the same two numbers cannot be reused at the same time. Find all the solutions that satisfy the condition, and guarantee that the solution will not be duplicated.

Code Implementation
``Import Java.util.arraylist;import java.util.arrays;import java.util.linkedlist;import java.util.List; Public classSolution { PublicList<list<integer>> foursum (int[] num, int target) {list<list<integer>> result =NewLinkedlist<> ();if(num = =NULL|| Num.length <4) {returnResult } arrays.sort (num);//Sort an array         for(int i =0; I < num.length-3; i++) {//First addend            if(I >0&& Num[i] = = Num[i-1]) {//First addend use non-repeatingContinue } for(Int j = num.length-1; J > i +2; j--) {//Fourth a addend                if(J < Num.length-1&& Num[j] = = Num[j +1]) {//Fourth Addend use non-repeatingContinue } int start = i +1;//A second addendIntEnd= J-1;//A third addendint n = target-num[i]-num[j]; while(Start <End) {if(Num[start] + num[End] = = N) {list<integer> four =NewArraylist<> (4);                        Four.add (Num[i]);                        Four.add (Num[start]); Four.add (num[End]);                        Four.add (Num[j]); Result.add (four); Do{start++; } while(start<End&& Num[start] = = Num[start-1]);//Guaranteed re-use second number not duplicated                         Do{End--; } while(Start <End&& num[End] = = num[End+1]);//Guaranteed re-use third number not duplicated}Else if(Num[start] + num[End] < N) { Do{start++; } while(start<End&& Num[start] = = Num[start-1]);//Guaranteed re-use second number not duplicated}Else{ Do{End--; } while(Start <End&& num[End] = = num[End+1]);//Guaranteed re-use third number not duplicated}                }            }        }returnResult }}``
Evaluation Results

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"Leetcode-Interview algorithm classic-java Implementation" "018-4sum (four numbers and)"

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