题解：3Sum Closest-优快云博客

本文链接：https://blog.youkuaiyun.com/WaverDn/article/details/70196523

探讨了如何寻找数组中三个整数之和最接近给定目标值的问题，并提供了一种有效的解决方案。通过先对数组进行排序，再利用双指针技巧遍历数组，实现了O(n^2)的时间复杂度。

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题目：

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).

解题思路：首先将该向量排序。之后，设置first, second, third 三个坐标，first自然从最小也就是0开始，second随其后，third从最后开始往前遍历。用一个大循环套一个小循环，大循环是first的递增，小循环中，如果检测到三个数的和比target大，那么third的数值减小，否则second增加，知道判断second > third时跳出小循环。就这样往复得到result.

代码如下：

int threeSumClosest(vector<int>& nums, int target) {
    sort(nums.begin(), nums.end());
    int result = nums[0] + nums[1] + nums[2];
    for(int first = 0; first < nums.size() - 2; first++) {
        int second = first + 1, third = nums.size() - 1;
        while(second < third) {
            int num = nums[first] + nums[second] + nums[third];
            if(abs(num - target) < abs(result - target))
				result = num;
            if(num < target)
				second++;
            else
				third--;
        }
    }
    return result;
}