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]
]

如果我们固定一个数，那么该问题就可以转化为2Sum问题。对于2Sum类型的问题，我们可以先对原始数组进行一个排序，然后用两个指针一个从数组首部开始往后扫描，一个从数组尾部往前扫描，途中如果两个指针所指的数的和为目标值时就可以记录下来，当两个指针相遇时，此算法执行结束。另外这道题有个要注意的地方，我们不能返回重复的数组，所以在扫描的过程中要防止重复。

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function (nums) {
    const findNums = function (nums, target, start, end) {
        const res = [];
        let i = start, j = end;
        while (i < j) {
            let sum = nums[i] + nums[j] + target;
            if (sum === 0) {
                let tmpArr = [target, nums[i], nums[j]];
                res.push(tmpArr.slice());
                i++;
                j--;
                while (i < j && nums[i - 1] === nums[i]) {
                    i++; // 去重
                }
                while (i < j && nums[j + 1] === nums[j]) {
                    j--; // 去重
                }
            }
            else if (sum < 0) {
                i++;
            }
            else {
                j--;
            }
        }
        return res;
    };
    let totalLen = nums.length;
    const result = [];
    if (totalLen >= 3) {
        nums.sort((a, b) => { return a - b; })
        for (let i = 0; i < totalLen - 2; i++) {
            if (i && nums[i] === nums[i - 1]) {
                continue;
            }
            let numArrs = findNums(nums, nums[i], i + 1, totalLen - 1);
            for (let j = 0; j < numArrs.length; j++) {
                result.push(numArrs[j]);
            }
        }
    }
    return result;
};