第11周作业2(LeetCode16)

1. 题目描述

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).

2. 解决思路

这道题与LeetCode上的15题很类似，但是这道题少了对重复三元组合的判断，所以说这道题可以说还简单了一些。解决问题的关键还是对数组下标的灵活利用，一句话概括就是：先对数组排序，然后左右（头尾）向中间靠拢。

3. 完整代码

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

int threeSumCloset(vector<int> &num, int target){
int ret = num[0] + num[1] + num[2];
int len = num.size();

sort(num.begin(), num.end());

for (int i = 0; i < len - 3; i++)
{
//第一个数：num[i]
int j = i + 1;    //第二个数
int k = len - 1;    //第三个数
while (j < k)
{
int sum = num[i] + num[j] + num[k];
if (abs(sum - target) < abs(ret - target))
ret = sum;
if (sum < target)
{
++j;
}
else if (sum > target)
{
--k;
}
else
{
++j;
--k;
}
}
}
return ret;
}

int main(){
vector<int> array;
int temp = 0;
int key = 0;
cout << "请输入整数数组的各元素：\n" << endl;
while (cin >> temp)
{
array.push_back(temp);
if (cin.peek() == '\n')
{
break;
}
}
cout << "请输入目标值：\n" << endl;
cin >> key;
int result = threeSumCloset(array , key);
cout << "最接近目标值的三数之和是：\n" << result << endl;
}