Leetcode Problem Solving report--3 Sum

title: 3 Numbers and questions
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)

Original title link address: https://leetcode.com/problems/3sum/
Analysis: Test instructions is from the given array of integers, find out three numbers and 0 of all possible combinations, the problem has two methods to solve:
1. Violence law: Through three for the loop to find all triplet of the given array and the case, and through the IF statement, will and 0 of the triplet record, and print out! The time complexity is 0 (n^3), inevitably tle, unable to accepted.
2. Pointer Shift method: The original array is sorted first, then left (l), the Right (R) (pointer) variable points to the first element of the given array, the 3 numbers and problems into 2 numbers and 2sum problem. Time complexity O (NLOGN).
the above method should pay attention to the last removal of the heavy operation!
Brute Force Java code (timeout cannot be accepted)

ImportJava.util.ArrayList;ImportJava.util.Arrays;ImportJava.util.Collection;ImportJava.util.HashSet;ImportJava.util.Set;ImportCom.sun.xml.internal.bind.v2.runtime.unmarshaller.XsiNilLoader.Array;ImportCom.sun.xml.internal.bind.v2.schemagen.xmlschema.List; Public  class threesum {    /** * @param args * *     Public Static void Main(string[] args) {//TODO auto-generated method stub        int[] s =New int[] {0,2,1,-3}; System.out.println ("A solution set is:"); Arraylist<arraylist<integer>> Listarray =NewArraylist<arraylist<integer>> (); Listarray = Threesum (s); for(inti =0; I < listarray.size ();        i++) {System.out.println (Listarray.get (i)); }    } Public StaticArraylist<arraylist<integer>>Threesum(int[] nums) {arraylist<arraylist<integer>> list =NewArraylist<arraylist<integer>> (); String string =""; arraylist<string> string1 =NewArraylist<string> ();int[] B =New int[3]; for(inti =0; i < Nums.length;i + +) { for(intj = i +1; J < Nums.length;j + +) { for(intK = j +1K < nums.length;k + +) {arraylist<integer> element =NewArraylist<integer> ();if(Nums[k] + nums[j] + nums[i] = =0) {string ="";                        Element.add (Nums[i]);                        Element.add (Nums[j]);                        Element.add (Nums[k]); b[0] = Nums[i]; b[1] = Nums[j]; b[2] = Nums[k];//avoid reduplicated ResultArrays.sort (b); for(intn =0N < B.length;n + +) {string = string + B[n]; }if(!string1.contains (String))                            {List.add (element);                        String1.add (string); }                    }                }            }        }returnList }}

End-to-end pointer shift method Java code (accepted)

     Public StaticArraylist<arraylist<integer>> Threesum (int[] nums) {arraylist<arraylist<integer>> list =NewArraylist<arraylist<integer>> (); Hashset<arraylist<integer>> ElementSet =NewHashset<arraylist<integer>> ();int sum=0; Arrays.sort (Nums); for(inti =0; I < nums.length-2; i++) {//directly Avoid reduplicated result            intleft = i +1;intright = Nums.length-1; while(Left < right) {sum= Nums[i] + nums[left] + nums[right];if(sum==0) {arraylist<integer> element =NewArraylist<integer> ();                    Element.add (Nums[i]);                    Element.add (Nums[left]); Element.add (Nums[right]);if(!elementset.contains (Element))                        {List.add (element);                    Elementset.add (Element);                    } left + +;                Right--; }Else if(sum<0) {left + +; }Else{Right--; }            }        }returnList }}

Test results:

 A Solution Set  is:[ -15, 1, +][ -15, 2,][ -15, 3, a][ -15, 4, one][ -15, 5, ten][ -15, 6, 9][ -15, 7, 8][ -14, 0, +][ -14, 1,][ -14, 2, a][ -14, 3, one][ -14, 4, ten][ -14, 5, 9][ -14, 6, 8][ -14, 7, 7][ -13,-1, +][ -13, 0,][ -13, 1, a][ -13, 2, one][ -13, 3, ten][ -13, 4, 9][ -13, 5, 8][ -13, 6, 7][ -12,-2, +][ -12,-1,][ -12, 0, a][ -12, 1, one][ -12, 2, ten][ -12, 3, 9][ -12, 4, 8][ -12, 5, 7][ -12, 6, 6][ -11, -3, +][ -11,-2,][ -11,-1,][ -11, 0, one][ -11, 1, ten][ -11, 2, 9][ -11, 3, 8][ -11, 4, 7][ -11, 5, 6][ -10, -4, +][ -10, -3,][ -10,-2,][ -10,-1, one][ -10, 0, ten][ -10, 1, 9][ -10, 2, 8][ -10, 3, 7][ -10, 4, 6][ -10, 5, 5][-9, -5, +][-9, -4,][-9, -3,][-9,-2, one][-9,-1, ten][-9, 0, 9][-9, 1, 8][-9, 2, 7][-9, 3, 6][-9, 4, 5][-8,-6, +][-8, -5,][-8, -4,][-8, -3, one][-8,-2, ten][-8,-1, 9][-8, 0, 8][-8, 1, 7][-8, 2, 6][-8, 3, 5][-8, 4, 4][-7,-7, +][-7,-6,][-7, -5,][-7, -4, one][-7, -3, ten][-7,-2, 9][-7,-1, 8][-7, 0, 7][-7, 1, 6][-7, 2, 5][-7, 3, 4][-6,-6,][-6, -5, one][-6, -4, ten][-6, -3, 9][-6,-2, 8][-6,-1, 7][-6, 0, 6][-6, 1, 5][-6, 2, 4][-6, 3, 3][ -5, -5, ten][ -5, -4, 9][ -5, -3, 8][ -5,-2, 7][ -5,-1, 6][ -5, 0, 5][ -5, 1, 4][ -5, 2, 3][ -4, -4, 8][ -4, -3, 7][ -4,-2, 6][ -4,-1, 5][ -4, 0, 4][ -4, 1, 3][ -4, 2, 2][ -3,-2, 5][ -3,-1, 4][ -3, 0, 3][ -3, 1, 2][-2,-2, 4][-2,-1, 3][-2, 0, 2][-2, 1, 1][-1,-1, 2][-1, 0, 1][0, 0, 0]

The relevant code is placed in the personal GITHUB:HTTPS://GITHUB.COM/GANNYEE/LEETCODE/TREE/MASTER/SRC

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. Personal GitHub code space: Https://github.com/gannyee

Leetcode Problem Solving report--3 Sum