LeetCode15. Sum of three

Ideas

Use double pointers to .

First, sort the array .

Three pointers , A pointer to the current element i, A left pointer left, A right pointer right.

Give Way i Traverse the array subscript at once ,left=i+1;right = nums.length-1;temp = nums[i]+nums[left]+nums[right]; When temp Greater than 0 when , will right--; When temp Less than 0 Just when left++; When temp==0 Just when left++,right--;

The traversal result does not exist set in ,set Can automatically remove duplicate elements . The final will be set To list that will do .

Code

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        int left = 0;
        int right = nums.length-1;
        List<List<Integer>> res = new ArrayList<>();
        Set<List<Integer>> t = new HashSet<>();
        int temp = 0;
        for(int i = 0;i<=right;i++){
            left = i+1;
            right = nums.length-1;
            while(right>left){
                temp = nums[i]+nums[left]+nums[right];
                if( temp== 0){
                    t.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    right--;
                    left++;
                }else if(temp>0){
                    right--;
                }else{
                    left++;
                }
            }
        }
        for(List l :t){
            res.add(l);
        }
        return res;
    }
}