3 Sum Closest

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

For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

题解1 - 排序 + 2 Sum + 两根指针 + 优化过滤

和 3 Sum 的思路接近，首先对原数组排序，随后将3 Sum 的题拆解为『1 Sum + 2 Sum』的题，对于 Closest 的题使用两根指针而不是哈希表的方法较为方便。对于有序数组来说，在查找 Cloest 的值时其实是有较大的优化空间的。

C++:

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) 
    {
        if (num.size() <= 3) return accumulate(num.begin(), num.end(), 0);
        sort (num.begin(), num.end());

        int result = 0, n = num.size(), temp;
        result = num[0] + num[1] + num[2];
        for (int i = 0; i < n - 2; ++i)
        {
            int j = i + 1, k = n - 1;
            while (j < k)
            {
                temp = num[i] + num[j] + num[k];

                if (abs(target - result) > abs(target - temp))
                    result = temp;
                if (result == target)
                    return result;
                ( temp > target ) ? --k : ++j;
            }
        }
        return result;
    }
};

源码分析

和前面3Sum解法相似，同理使用i,j,k三个指针进行循环。
区别在于3sum中的target为0，这里新增一个变量用于比较哪组数据与target更为相近

复杂度分析

时间复杂度同理为O(n^2)运行时间 16ms