題目：

Given array nums = [-1, 2, 1, -4], and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

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).

思路：和上一題類似，要保證和target的差 最小，那麼定義 一個變數diff儲存他們之間的差。首先排序，然後固定一個位置，剩下的兩個數left從左往右，right從又往左，每確定兩個數，我們求出此三數之和，然後算和給定值的差的絕對值存在newDiff中，然後和diff比較並更新diff和結果closest即可

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

參考：http://www.cnblogs.com/grandyang/p/4510984.html