# Simple Java Solution - Without using Hash/Set

• ``````public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> list = new ArrayList<List<Integer>>();
for(int i = 0; i < nums.length-2; i++) {
if(i > 0 && (nums[i] == nums[i-1])) continue; // avoid duplicates
for(int j = i+1, k = nums.length-1; j<k;) {
if(nums[i] + nums[j] + nums[k] == 0) {
j++;k--;
while((j < k) && (nums[j] == nums[j-1]))j++;// avoid duplicates
while((j < k) && (nums[k] == nums[k+1]))k--;// avoid duplicates
}else if(nums[i] + nums[j] + nums[k] > 0) k--;
else j++;
}
}
return list;
}``````

• The idea here is to sort the array first. Then for each element(`i`), have two pointers at the beginning(`j`), and end(`k`). Check if 3sum == 0.

To avoid duplicates, if any of the next number is same, just skip it. That is, if `nums[i]` is same as previous, skip it. Similarly for `nums[j]` and `nums[k]`

• it will take less time if we write `sum=nums[i] + nums[j] + nums[k]` at first.

• that's a great thinking!

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