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).

翻譯
給定一個包括 n 個整數的陣列 nums 和 一個目標值 target。找出 nums 中的三個整數，使得它們的和與 target 最接近。返回這三個數的和。假定每組輸入只存在唯一答案。

例如，給定陣列 nums = [-1，2，1，-4], 和 target = 1.
與 target 最接近的三個數的和為 2. (-1 + 2 + 1 = 2).

分析
這題其實比3Sum還要簡單，方法相同，由於只有一組解，所以不用考慮重複。

c++實現

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