leetcode(15)3sum

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: 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]
]

因爲本人比較菜，所以一開始想的當然是N3的遍歷，但是一個medium的題目肯定不能這麼簡單，所以這題就理解爲成a+b = -c的題目。然後可以用雙指針做，遍歷num[]，0-num[i]作爲-c，left指針從i+1開始向右遍歷，right從右開始向左遍歷。

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        //先排序
        Arrays.sort(nums);
        //避免重複，將獲得的list先放入set中
         Set<List> sets = new HashSet<>();
        for(int i=0;i<nums.length;i++){
         //如果num[i]>0就出現不了=0的情況所以break
            if (nums[i]>0)break;
            int target = 0-nums[i];
            int j = i+1;int k = nums.length-1;//定義左右指針
            while (j<k){
                if (nums[j]+nums[k] == target){
                    List<Integer> set = new ArrayList<>();
                    set.add(nums[i]);set.add(nums[j]);set.add(nums[k]);
                    sets.add(set);
                    j++;k--;
                }
                //如果左右指針的數的sum小於target則從用更大的比較，所以j++，同理，如果小於target，則k--;
                else if (nums[j]+nums[k]<target){
                    j++;
                }
                else {
                    k--;
                }
            }
        }
        List<List<Integer>> lists = new ArrayList<>();
        for (List list: sets){
            lists.add(list);
        }
        return lists;
    }
}