Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:

Given array nums = [-1, 2, 1, -4], and target = 1.The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

这个题目跟很类似，但是有target，然后要最接近的，所以还是用two pointers去得到每个num 之后的pair，使得他们的sum最接近target，然后取所有sum中最接近的sum.

class Solution:    def 3sumClosest(self, nums, target):        ans, dif, n = None, None, len(nums)        nums.sort()        for i in range(n - 2):            total = nums[i] + self.helper(nums, i + 1, n - 1, target - nums[i])            localDif = abs(total - target)            if dif is None or localDif < dif:                    dif, ans = localDif, total        return ans    def helper(self, nums, start, end, target):        ans, dif = None, None        while start < end:            total = nums[start] + nums[end]            if total == target:                return total            elif total < target:                localDif = target - total                start += 1            elif total > target:                localDif = total - target                end -= 1            if dif is None or localDif < dif:                dif, ans = localDif, total        return ans