每日算法之十五：threesumClosset

标签：近似三元组

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).

<span style="font-size:14px;">class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        
                sort(num.begin(), num.end());
         
        int ret;
         bool first = true;
         
         for(int i = 0; i < num.size(); i++)
         {
             int j = i + 1;
             int k = num.size() - 1;
             
             while(j < k)
             {
                 int sum = num[i] + num[j] + num[k];
                 if (first)
                 {
                     ret = sum;
                     first = false;
                 }
                 else
                 {
                     if (abs(sum - target) < abs(ret - target))
                         ret = sum;                                         
                 }
                 
                 if (ret == target)
                     return ret;
                 
                 if (sum > target)
                     k--;
                 else
                     j++;
             }
         }
         
         return ret;
     }
    };
</span>

每日算法之十五：threesumClosset,布布扣,bubuko.com

每日算法之十五：threesumClosset

标签：近似三元组

原文地址：http://blog.csdn.net/yapian8/article/details/26749455