16. 3Sum Closest

思路：
先任意取三元素和res；记录和目标的差值diff
每次取一个数，在后续列表中找2个元素，满足三数之和最接近target，通过判断三数之和和目标的差值和diff大小，小于diff则将三数之和和差值赋值给res和diff,然后再分析和小于移动左指针，大于移动右指针，直到遍历找到最接近的数

Python：

class Solution(object):    def threeSumClosest(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: int        """        if len(nums) < 3:            return []        nums.sort()           #先任意取三元素和triple；记录和目标的差值diff                      res = nums[0] + nums[1] + nums[2]          diff = abs(target - res)        for i in range(0,len(nums)-2):            if nums[i] != nums[i-1] or i == 0:                l = i + 1                r = len(nums) - 1                while l < r:                    s = nums[i] + nums[l] + nums[r]                    d = abs(target - s)   #每次都计算差值                    if d < diff:                        diff = d                        res = s                    if s == target:                        return target                    elif s < target:                        l += 1                    else:                        r -= 1        return res      

C++：

class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {        if (nums.size() < 3){            return 0;          }        sort(nums.begin(),nums.end());        int res = nums[0] + nums[1] + nums[2];        int diff = abs(target - res);        for (int i = 0; i < nums.size()-2;++i){            if (nums[i] == nums[i-1] && i > 0){                continue;            }            int left = i + 1;            int right = nums.size() - 1;            while(left < right){                int s = nums[i] + nums[left] + nums[right];                int d = abs(target - s);                if (d < diff){                    diff = d;                    res = s;                }                if(s == target){                    return target;                                 }                else if (s < target){                    left ++;                }                else{                    right --;                }            }        }        return res;    }};