3sum apps

1.Given an array S of n integers, is there elements a, b, c in S such That a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.Note: Elements in a triplet (a,b,C) must is in non-descending order. (ie, a ≤ b ≤ c) The solution set must not contain duplicate triplets. For example, given array S = {-1 0 1 2-1-4}, A solution set is: ( -1, 0, 1) (-1,-1, 2)2.This problem extended with 2sum problems, and then can continue to extend 4sum.The con

) + returnsum; A if(difmin) { atmin=dif; -result=sum; - } - //if sum is already greater than target, then it doesn't make sense to continue looking for the third digit position. (also to make the best use of array ordering) - if(Nums[index]>0 sum>target) - Break; in //the element has a repeating condition - while(Index+1Num

Sumsets Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9997 Accepted: 2736 DescriptionGiven S, a set of integers, find the largest d such that A + B + c = d where a, B, C, and D is distinct elements of S.InputSeveral S, each consisting of a line containing an integer 1 OutputFor each S, a single line containing d, or a single line containing "no solution".Sample Input52 3 5 7 1252 16 64 256 10240Sample Output12NO Solutio

, skip; } //Do 2Sum intHead = i + 1; intTail = num.length-1; while(Head tail) { if(Head > i + 1 num[head] = = Num[head-1]) {head++; Continue;//If current value is already tested as the second, skip; } if(Tail ]) {tail--; Continue;//similar test for the third element } intsum = Num[i] + Num[head] +Num[tail]; if(Sum = =target) {Solution.add (NewA

QuestionGiven an array of n integers nums and a target, find the number of index triplets with i, j, k that satisfy the condition nums[i] + nums[j] + nums[k] .For example, given nums = [-2, 0, 1, 3] , and target = 2.Return 2. Because there is triplets which sums is less than 2:[-2, 0, 1] [-2, 0, 3]SolutionSince this problem does not require a heavy weight, we do not consider repetition.Although the topic mentions index, we find that the number is returned. So it is possible to sort the arr

Translation给定一个有n个整数的数组S，是否存在三个元素a,b,c使得a+b+c=0?找出该数组中所有不重复的3个数，它们的和为0。备注：这三个元素必须是从小到大进行排序。结果中不能有重复的3个数。例如，给定数组S={-1012 -14}，一个结果集为：(-101)(-1, -12)OriginalGiven anArray S ofn integers, is there elementsa, B, CinchS such thata+ B + c =0? Find All unique tripletsinch theArray which gives the sum of Zero. Note:elementsinch aTriplet (a, b,c) must beinchnon-DescendingOrder. (ie,a≤B≤C) The solutionSetMust notContain duplicate triplets. For example, given array S = {-1 0 1 2-1-4},a SolutionSetis: (-1

Problem:Given an array S of n integers, find three integers in S such, the sum was closest to a Given number, target. Return the sum of the three integers. You may assume this each input would has exactly one solution. For example, given array S = {-1 2 1-4}, and target = 1. The sum is closest to the target is 2. (-1 + 2 + 1 = 2).Given an array, the number of three numbers is closest to the target, and the result is nearly the default uniqueThinking:(1) is not very familiar, this problem i

TopicGiven an array S of n integers, is there elements a, B, C in S such that A + B + c = 0? Find all unique triplets in the array which gives the sum of zero.Note:Elements in a triplet (A,B,C) must is in non-descending order. (ie, a≤b≤c)The solution set must not contain duplicate triplets.For example, given array S = {-1 0 1 2-1-4},A solution set is:(-1, 0, 1)(-1, -1, 2)AnalysisThe sequence of three of the sum of 0 in a given integer sequence is calculated!I used a violent solution, three-layer

Given an array S of n integers, is there elements a, b, c in S such That a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.Note: Elements in a triplet (a,b,C) must is in non-descending order. (ie, a ≤ b ≤ c) The solution set must not contain duplicate triplets. For example, given array S = {-1 0 1 2-1-4}, A solution set is: ( -1, 0, 1) (-1,-1, 2)Https://leetcode.com/discuss/23638/concise-o-n-2-java-solutionHttp://www.cnblogs.com/tenos

? This does not have to worry about 0 of the problem.3 Code: PublicListint[] nums) {Arrays.sort (nums); intLen =nums.length; ListNewArraylist(); for(inti = 0; i ){ if(i = = 0 | | (i > 0 nums[i]!=nums[i-1])){ inthi = len-1; intLo = i + 1; intsum = 0-Nums[i]; while(Lo hi) { if(Nums[lo] + nums[hi] = =sum) {Answerlist.add (Arrays.aslist (Nums[i],nums[lo],nums[hi])); while(Lo ; while(Lo ; Lo++; Hi--; }Else if(Nums

This problem and 015 3 SUM basic idea is the clamping force of Thought complexity O (n*n)classSolution:#@param {integer[]} nums #@param {integer} target #@return {integer} defthreesumclosest (self, Nums, target): Length=Len (nums) Nums=Sorted (nums) ans= SUM (nums[0:3]) I=0 whileI : J= i + 1k= Length-1 whileJ k:s= Nums[i] + nums[j] +Nums[k]ifs = =Target:returnTargetElse: ifABS (Ans-target) > abs (S-target): ans=sifs target:j+ = 1Else: K-= 1I+ = 1returnAns016

figured it out yet. Ten //while (i>0nums[i]==nums[i-1]) ++i; Pay attention to comparing two ways of writing One if(i>0nums[i]==nums[i-1])Continue; Aj=i+1; -k=size-1; - while(jk) { the if(j>i+1nums[j]==nums[j-1]){ -++J; - Continue; - } + if(Nums[j]+nums[k]>-nums[i])--K; - Else if(Nums[j]+nums[k]J; + Else{ Atemp[0]=Nums[i]; attemp[1]=Nums[j]; -temp[2]=Nums[k]; -

the third variable are synchronized to increase or decrease, and to determine whether there are duplicate numbers in the process of self-increment; Recursively until the condition of the first variable expires.Code (c + +):classSolution { Public: vectorvectorint> >Threesum ( vectorint>nums) { vectorvectorint> >Result Sort (Nums.begin (), Nums.end ()); for(inti =0; I if(I >0 Nums[i] = = nums[i-1])Continue; Threenumber (nums, result, i); }returnResult }//return vector voidThreenum

); Private: intresult, finalresult, Gap =99999999, Gap1 =99999999;};intSolution::threesumclosest (vectorint> Nums,inttarget) { //int result,gap=99999999;sort (Nums.begin (), Nums.end ()); intLen =nums.size (); for(inti =0; I 2; i++) { //cout intRes1 = Findclosest (Nums, I, Len-1, Target); if(Res1 gap1) Gap1=res1; Finalresult=result; } //cout //return gap1; returnFinalresult;}intSolution::findclosest (vectorint> Nums,int Base,intEndinttarget) { //cout intBegi

Code:1#include 2#include 3#include 4 5 using namespacestd;6 7 intThreesumclosest (vectorint> num,inttarget) {8 sort (Num.begin (), Num.end ());9 intL =num.size ();Ten intMin =Int_max; One intBeauty; A for(inti =0; I 2; i++) - { - intL = i +1; the intr = L-1; - intsum = num[l]+num[r]+Num[i]; - while(L R) - { +sum = Num[l] + num[r] +Num[i]; - if(ABS (Target-sum) min) + { Amin = ABS (Target-sum); atBeauty =sum; -

The algorithm slag, the reality basic reference or completely copies [Dai Fangqin ([email protected])] Great God's leetcode problem, here only for brush title records.1 classSolution {2 Public:3 intThreesumclosest (vectorint> num,inttarget) {4 intresult =0;5 intMin_gap =Int_max;6 7Sort (Num.begin (), Num.end ());//Sort8 9Auto last =num.end ();Ten for(Auto a = Num.begin (); a 2); ++a) { OneAuto B =next (a); AAuto C =prev (last); - while(b c) { -

;0) {Big--; } Else if(Sum 0) {Mid++; } Else{vectorint>v; V.push_back (Num[small]); V.push_back (Num[mid]); V.push_back (Num[big]); Ans.push_back (v); Do{mid++;} while(Mid 1] = = Num[mid]);//Attention!! Do{big--;} while(Mid 1] = = Num[big]);//Attention!! } } Do{small++;} while(Small 1 Num[small-1] = = Num[small]);//Attention!!

Given an array S of n integers, find three integers in S such so the sum is closest to a give n number, target. Return the sum of the three integers. You may assume this each input would has exactly one solution. For example, given array S = {-1 2 1-4}, and target = 1. The sum is closest to the target is 2. (-1 + 2 + 1 = 2).Has you met this question in a real interview? Solution:1 Public classSolution {2 Public intThreesumclosest (int[] num,inttarget) {3 intLen =num.length;4

TopicGiven an array S of n integers, find three integers in S such so the sum is closest to a give n number, target. Return the sum of the three integers. You may assume this each input would has exactly one solution. For example, given array S = {-1 2 1-4}, and target = 1. The sum is closest to the target is 2. (-1 + 2 + 1 = 2).AnswerThe topic requires finding the sum of the three elements in the array and closest to the target value, and then returning the values of the three numbers.Fir

TopicGiven an array S of n integers, is there elements a, b, c in S such That a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.note: Elements in a triplet (a,b,C) must is in non-descending order. (ie, a ≤ b ≤ c) The solution set must not contain duplicate triplets. For example, given array S = {-1 0 1 2-1-4}, A solution set is: ( -1, 0, 1) (-1,-1, 2)Codepublic class Solution {public listcode Download: Https://github.com/jimenbi