3Sum Closest : https://leetcode.com/problems/3sum-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).
解析:
根据上一题 3 Sum 中的夹逼算法,可解此题。
不同的是无需考虑是否有重复的三个元素组,因为无需记录最接近target的元素组成
#include <vector>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
using namespace std;
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int result = nums[0]+nums[1]+nums[2];//初始化
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size()-2; i++) {
int j = i+1;
int k = nums.size()-1;
int sum;
while (j < k) {
sum = nums[i]+nums[j]+nums[k];
if (abs(sum-target) < abs(result-target))
result = sum;
if (sum < target)
j++; //不需检测nums[j]==nums[j-1],可以重复,不需保存满足条件的元素
else if (sum > target)
k--;
else
return target;
}
//注:不可将第18、19行放在此处,因为此处的终止条件是j == k
}
return result;
}
};
int main()
{
int num[] = {87,6,-100,-19,10,-8,-58,56};
vector<int> nums(num, num+sizeof(num)/sizeof(num[0]));
Solution sol;
cout << sol.threeSumClosest(nums, 10) << endl; //87 -19 -58 : 10
cout << sol.threeSumClosest(nums, 9) << endl; //87 -19 -58 : 10
cout << sol.threeSumClosest(nums, 8) << endl;
getchar();
}
原文地址:http://blog.csdn.net/quzhongxin/article/details/46010781