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)
Have you met this question in a real interview?
Analysis:
Solution:
1 public class Solution { 2 public List<List<Integer>> threeSum(int[] num) { 3 List<List<Integer>> resList = new ArrayList<List<Integer>>(); 4 if (num.length<3) return resList; 5 6 Arrays.sort(num); 7 Map<Integer,Integer> numMap = new HashMap<Integer,Integer>(); 8 for (int i=0;i<num.length;i++) 9 if (numMap.containsKey(num[i])) 10 numMap.put(num[i],numMap.get(num[i])+1); 11 else numMap.put(num[i],1); 12 13 int first = 0; 14 int second = -1; 15 while (first<num.length-2){ 16 numMap.put(num[first],numMap.get(num[first])-1); 17 second = first+1; 18 while (second<num.length-1){ 19 numMap.put(num[second],numMap.get(num[second])-1); 20 int left = 0-num[first]-num[second]; 21 if (left>=num[second] && numMap.containsKey(left) && numMap.get(left)>0){ 22 List<Integer> temp = new LinkedList<Integer>(); 23 temp.add(num[first]); 24 temp.add(num[second]); 25 temp.add(left); 26 resList.add(temp); 27 } 28 numMap.put(num[second],numMap.get(num[second])+1); 29 if (left<num[second]) break; 30 while (second< num.length-1 && num[second]==num[second+1]) second++; 31 second++; 32 } 33 numMap.put(num[first],numMap.get(num[first])+1); 34 while (first<num.length-2 && num[first]==num[first+1]) first++; 35 first++; 36 37 } 38 39 return resList; 40 } 41 }
NOTE: There is a way to solve this problem without using hashmap. Practice it!
Solution2: Two pointers.
The key point here is that:
if p1+p2+p3==0, then p2++ and p3--.
1 public class Solution { 2 public List<List<Integer>> threeSum(int[] num) { 3 List<List<Integer>> res = new ArrayList<List<Integer>>(); 4 if (num.length<3) return res; 5 Arrays.sort(num); 6 7 8 9 //Find each solution. 10 int p1 = 0; 11 int p2,p3; 12 while (p1<num.length-2 && num[p1]<=0){ 13 p2 = p1+1; 14 p3 = num.length-1; 15 while (p2<p3 && -num[p1]-num[p2]>=num[p2]){ 16 int sum = num[p1]+num[p2]+num[p3]; 17 if (sum==0){ 18 List<Integer> sol = new ArrayList<Integer>(); 19 sol.add(num[p1]); 20 sol.add(num[p2]); 21 sol.add(num[p3]); 22 res.add(sol); 23 } 24 25 if (sum<=0){ 26 p2++; 27 while (p2<p3 && num[p2]==num[p2-1]) p2++; 28 } 29 30 if (sum>=0){ 31 p3--; 32 while (p2<p3 && num[p3]==num[p3+1]) p3--; 33 } 34 } 35 p1++; 36 while (p1<num.length-2 && num[p1]==num[p1-1]) p1++; 37 } 38 39 return res; 40 41 } 42 }