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.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]

tag: array, two pointers

method 1

可以将3sum问题转换为2sum问题，然后再用2sum的办法解决。这里使用两个指针的方法

先对数组排序，这是为了让指针更好地移动，让指针有方向

遍历每一个索引i上的数nums[i]，然后讨论i+1~length-1这个范围内是否有数的和等于nums[i]的负数

public List<List<Integer>> threeSum(int[] nums) {
    Arrays.sort(nums);
    List<List<Integer>> res = new ArrayList<>();

    for (int i = 0; i < nums.length-2; i++) {

        if (i == 0 || (i > 0 && nums[i] != nums[i-1])){
            int low = i+1, high = nums.length-1, sum = -nums[i];
            while (low < high){
                if (nums[low] + nums[high] == sum){
                    res.add(Arrays.asList(nums[i],nums[low],nums[high]));
                    while (low < high && nums[low] == nums[low+1]) low++;
                    while (low < high && nums[high] == nums[high-1]) high--;
                    low++;high--;
                }else if (nums[low] + nums[high] < sum) low++;
                else high--;
            }
        }
    }

    return res;
}

summary :

1. 在使用两根指针的时候，考虑一下是否需要排序，让指针的移动更具有方向性
2. 当处理去除重复性问题时，既可以使用set自动去重，也可以进行排序，当数重复的时候便跳过