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).
3 sum 的变形,这里不需要考虑重复而已
class Solution { public: int threeSumClosest(vector<int> &num, int target) { // Start typing your C/C++ solution below // DO NOT write int main() function sort(num.begin(), num.end()); int len = num.size(); if(len < 3) return 0; int res = num[0] + num[1] + num[2]; for(int i = 0; i <= len -3; ++i){ int low = i+1; int high = len -1; while(low < high){ int sum = num[i] + num[low] + num[high]; if(sum == target) return target; else if(sum < target){ ++low; if(abs(sum - target) < abs(res - target)) res = sum; }else{ --high; if(abs(sum - target) < abs(res - target)) res = sum; } } } return res; } };