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).
解题思路:
与Three Sum的做法类似,首先枚举中间数字的位置,然后找寻和离target减去中间数字最近的
点即可(此处需用绝对值).时间复杂度为O(N^2).
解题代码:
class Solution { public: int threeSumClosest(vector<int> &num, int target) { sort(num.begin(),num.end()); unsigned n = num.size() ; long long res = abs(num[0] + num[1] + num[2] - target) , sum = num[0] + num[1] + num[2] ; for(unsigned i=1;i < n - 1 ;++i) { long long sum1 = target - num[i] ; unsigned j = 0 , k = n - 1 ; while( j < i && k > i ) { long long tmp = num[j] + num[k] ; if(res > min(res,abs(sum1-tmp))) { res = min(res,abs(sum1-tmp)); sum = num[i] + tmp; } if(sum1 == tmp) return sum; tmp > sum1 ? --k : ++j ; } } return sum; } };
LeetCode:3Sum Closest,布布扣,bubuko.com
原文地址:http://blog.csdn.net/dream_you_to_life/article/details/26763461