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).
本文是3sum问题的变更,是寻找到最接近目标值的三数组合,利用diff存放与目标最小的差距,如果diff=0,即可输出。
参考的3sum问题http://www.cnblogs.com/Scorpio989/p/4440236.html
时间:15ms。代码如下:
class Solution { public: int threeSumClosest(vector<int>& nums, int target) { int besttarget = 0, diff = INT_MAX; sort(nums.begin(), nums.end()); for (int i = 0; i < nums.size() - 2; i++){ int j = i + 1, n = nums.size() - 1; if (i>0 && nums[i] == nums[i - 1]) continue; while (j < n){ if (j>i + 1 && nums[j] == nums[j - 1]){ j++; continue; } if (n<nums.size() - 1 && nums[n] == nums[n + 1]){ n--; continue; } int sum = nums[i] + nums[j] + nums[n]; if (abs(sum - target) < diff){ diff = abs(sum - target); besttarget = sum; } if (sum == target) return besttarget; else if (sum>target) n--; else j++; } } return besttarget; } };