Description：
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).
思路：
本题要求的是找到三个数，满足三数之和与target的距离最近，思路与3sum大体相同，仍然是固定一个数，用两指针双向往中间走，记录并维护当前三数之和与target的最小距离的差值mindiff，实现代码如下：
class Solution(object):
def threeSumClosest(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
nums.sort()
mindiff= float("inf")
n = len(nums)
for i in range(n - 2):
if i > 0 and nums[i] == nums[i - 1]: continue
low = i + 1
high = n - 1
while low < high:
cursum = nums[low] + nums[high] + nums[i]
if cursum == target:
while low < high and nums[low] == nums[low + 1]:
low = low + 1
while low < high and nums[high] == nums[high - 1]:
high = high - 1
high = high - 1
low = low + 1
elif cursum > target:
high = high - 1
else:
low = low + 1
# 比较mindiff和当前计算差值的绝对值
if abs(mindiff) > abs(target - cursum):
# 每次记录的都是target - 当前三数之和
mindiff = target - cursum
return target - mindiff