标签：ber turn public ati put ref ase hit pre
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).1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 | class Solution {public: int threeSumClosest(vector<int>& nums, int target) { int len = nums.size(); if (len < 3) return 0; sort(nums.begin(), nums.end()); int sum = 0; int leastdif = INT_MAX; for (int i = 0; i < len - 2; i++) { int l = i + 1, r = len - 1; while (l < r) { int dif = fabs(nums[i] + nums[l] + nums[r] - target); if (dif < leastdif) { leastdif = dif; sum = nums[i] + nums[l] + nums[r]; } else if (nums[i] + nums[l] + nums[r] < target) { l++; } else { r--; } } } return sum; }}; |
标签：ber turn public ati put ref ase hit pre
原文地址：http://www.cnblogs.com/zhxshseu/p/af219256eafc904b6acab29753157811.html
