3Sum (LeetCode Array)

Question: 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)

Idea: First sort the array. Then use three pointers (i,j,k) to denote the triplet (a,b,c). Fix i, use j,k to maintain a sliding window. If the current sum is 0, add to result; if the current sum <0, move the left pointer forward, if the current sum>0, move the right pointer backward. Skip the continuously same numbers to avoid duplication.

Time: O(n^2) Space: O(1)

Code: 

 public class Solution {  
   public List<List<Integer>> threeSum(int[] num) {  
     List<List<Integer>> result=new ArrayList<List<Integer>>();  
     if(num==null||num.length<3)  
     {  
       return result;  
     }  
     Arrays.sort(num);  
     for(int i=0;i<num.length-2;i++)  
     {  
       if(i>0&&num[i]==num[i-1])  
         continue;  
       int j=i+1;  
       int k=num.length-1;  
       while(j<k)  
       {  
         int curSum=num[i]+num[j]+num[k];  
         if(curSum<=0)  
         {  
           if(curSum==0)  
             result.add(Arrays.asList(num[i],num[j],num[k]));  
           j++;  
           while(j<k&&num[j]==num[j-1])  
             j++;  
         }  
         else  
         {  
           k--;  
           while(j<k&&num[k]==num[k+1])  
             k--;  
         }  
       }  
     }  
     return result;  
   }  
 }