15. 三数之和 3Sum

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Notice that the solution set must not contain duplicate triplets.

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

方法：排序加双指针

先对数组进行排序，对于重复的数据，不再处理

当我们需要枚举数组中的两个元素时，如果我们发现随着第一个元素的递增，第二个元素是递减的，那么就可以使用双指针的方法

所以三重循环，保持第二重循环不变，而将第三重循环变成一个从数组最右端开始向左移动的指针

public List<List<Integer>> threeSum(int[] nums) {
        int n = nums.length;
        Arrays.sort(nums);
        List<List<Integer>> ans = new ArrayList<>();
        for (int first = 0; first < n; first++){
            if(first > 0 && nums[first] == nums[first -1]){
                continue;
            }
            int third = n - 1;
            int target = -nums[first];
            for (int second = first + 1; second < n; second++){
                if(second > first + 1 && nums[second] == nums[second - 1]){
                    continue;
                }
                while (second < third && nums[second] + nums[third] > target){
                    third--;
                }
                if (third == second) break;
                if(nums[second] + nums[third] == target){
                    List<Integer> list = new ArrayList<>();
                    list.add(nums[first]);
                    list.add(nums[second]);
                    list.add(nums[third]);
                    ans.add(list);
                }
            }
        }
        return ans;
    }