3Sum-LeetCode

Given an array S of n integers, are 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 be 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)

Solutions:

(1) Use Set to eliminate possible duplicates

(2) Start from the middle of the array and move to two ends to have O(N^2)

(3) Similar to 2Sum problem

Code:

import java.util.ArrayList;import java.util.Arrays;import java.util.HashSet;import java.util.List;import java.util.Set;import java.util.Vector;public class ThreeSum {public static void main(String args[]){int[] s = {0,0,0,0,0,0};Arrays.sort(s);int first = 0;int size = s.length;int last = size-1;HashSet<Vector<Integer>> triplets = new HashSet<Vector<Integer>>();for(int i = 0; i < s.length; i++)while(last!=first&&last!=i&&first!=i){if(s[last]+s[first]+s[i]==0){Integer[] arr = {};Vector<Integer> triplet = new Vector<Integer>(Arrays.asList(arr));  triplet.add(s[first]);  triplet.add(s[i]);  triplet.add(s[last]);  triplets.add(triplet);first++;}else if(s[last]+s[first]+s[i]<0)first++;else if(s[last]+s[first]+s[i]>0)last--;}ArrayList<Vector<Integer>> list = new ArrayList<Vector<Integer>>(triplets);System.out.println(list);}}

Others Better Solution:

public class Solution {public ArrayList<ArrayList<Integer>> threeSum(int[] num) {ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if (num.length < 3)return result; // sort arrayArrays.sort(num); for (int i = 0; i < num.length - 2; i++) {// //avoid duplicate solutionsif (i == 0 || num[i] > num[i - 1]) { int negate = -num[i]; int start = i + 1;int end = num.length - 1; while (start < end) {//case 1if (num[start] + num[end] == negate) {ArrayList<Integer> temp = new ArrayList<Integer>();temp.add(num[i]);temp.add(num[start]);temp.add(num[end]); result.add(temp);start++;end--;//avoid duplicate solutionswhile (start < end && num[end] == num[end + 1])end--; while (start < end && num[start] == num[start - 1])start++;//case 2} else if (num[start] + num[end] < negate) {start++;//case 3} else {end--;}} }} return result;}}