# Clean Java O(n^2) solution using two pointers

• ``````public class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(nums==null || nums.length<3) return result;
Arrays.sort(nums);
for(int i=0;i<nums.length;i++){
int low = i+1;
int high = nums.length -1;
while(low<high){
if(nums[i]+nums[low]+nums[high]==0){
while(i+1<nums.length && nums[i+1]==nums[i]) i++;
while(low+1<nums.length && nums[low]==nums[low+1]) low++;
while(high-1>=0 && nums[high]==nums[high-1]) high--;
low++;
high--;
} else if(nums[i]+nums[low]+nums[high]>0) high--;
else low++;
}
}
return result;
}
}
``````

The inner three while loops ensure that if there are duplicate numbers, only one of the duplicate number is added to the solution.

Suppose the array is [2, 2, 5, 9, -7]

When i = 0, low = 2, high = 4, the sum would be 0 and hence the solution would be added to the list.
Now if we don't have these while loops and we let the for loop increment i to 1, we will once again get the same solution since 2 appears twice in the array.

• why you use this three while loops ? can plzz explain