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).
class Solution { public: int threeSumClosest(vector<int> &num, int target) { sort(num.begin(),num.end()); int base = 0,left = 1,right = num.size() - 1; int minSum = num[base] + num[left] + num[right];int mindistance = abs(minSum - target); for(base = 0;base< num.size();base++) { left = base + 1; right = num.size() -1; while(left < right) { int sum = num[base] + num[left] + num[right]; int dis = sum - target ; if(dis > 0) { right--; }else if(dis <0) { left++; }else { return sum; } if(abs(dis) < mindistance) { minSum = sum; mindistance = abs(dis); } } } return minSum; } };
leecode -- 3sum Closet,布布扣,bubuko.com
原文：http://www.cnblogs.com/berkeleysong/p/3756580.html