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]
]

public List<List<Integer>> threeSum(int[] nums){
        int n = nums.length;
        Arrays.sort(nums);
        List<List<Integer>> lists = new ArrayList<>();
        for (int i=0;i<n-2;i++){
            if (i>0&&nums[i]==nums[i-1]){//假如有重复的元素跳过
                continue;
            }
            int target = 0-nums[i];
            int head = i+1;
            int tail = n-1;
            while (head<tail){
                int tmp = nums[head]+nums[tail];
                if (tmp>target){
                    tail--;
                }
                else if(tmp<target){
                    head++;
                }
                else{
                    lists.add(Arrays.asList(nums[i],nums[head],nums[tail]));
                    //缩减范围，避免重复
                    int k=head+1;
                    while (k<tail&&nums[k]==nums[k-1]) k++;
                    head = k;

                    k=tail-1;
                    while (k>head&&nums[k]==nums[k+1]) k--;
                    tail = k;
                }
            }
        }
        return lists;
    }

原先的想法是用Hashmap，像做TwoSum 一样做，发现在处理重复元素时比较复杂。最终在网上查到TwoSum可以用排序后，首尾值与target比较，前后值比较缩减范围避免重复元素。