16. 3Sum Closest**

16. 3Sum Closest**

​​https://leetcode.com/problems/3sum-closest/description/​​

题目描述

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:

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)​​.

给定一个整数数组 S, 在其中找到 3 个整数使得它们的和接近一个给定的 target. 返回这 3 个数的和.

解题思路

思路: 类似于 ​​15. 3Sum​​​. 为了得到最接近 ​​target​​​ 的值, 要计算使 ​​std::abs(target - sum)​​​ 最小的那个 ​​sum​​​. 为了得到这些 ​​sum​​​, 和 ​​15. 3Sum​​​ 一样, 首先对数组进行排序, 然后固定 ​​nums[i]​​​, 再对后面的内容进行 ​​2Sum​​.

C++ 实现 1

代码中使用 ​​res​​​ 来记录每次找到的最小的三元组 ​​a + b + c​​ 之和.

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

        int res = nums[0] + nums[1] + nums[nums.size() - 1];
        std::sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size() - 2; ++i) {
            int lo = i + 1, hi = nums.size() - 1;
            while (lo < hi) {
                int sum = nums[lo] + nums[hi] + nums[i];
                if (sum == target)
                    return sum;
                if (sum > target)
                    hi --;
                else {
                    lo ++;
                }
                if (std::abs(target - sum) < std::abs(target - res))
                    res = sum;
            }
        }
        return res;
    }
};