[LeetCode] 3Sum Closest

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).

分析：和3Sum一样，不能用O（n^3）遍历，需要有一些可以简化的步骤，三层遍历肯定Time Limited Exceeded！

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        int n = num.size();
        map<int,int> diff;//key是target和sum的差值的绝对值，value是sum值
        sort(num.begin(),num.end());
        int sum ;
        for(int i=1;i<n-1;i++){

               int l = 0, r= n-1;
               while(l<i && r>i){
               if(l>1 && num[l] == num[l-1]){
                  l++;
                  continue;
               }
               if(r<n-1 && num[r] == num[r+1]){
                  r--;
                  continue;
               }
               sum = num[l]+num[i]+num[r];
               if(sum == target)
                   return sum;
               else {
                   int abs   = sum-target>0 ? sum-target :target-sum;
                   diff[abs] = sum;
                   if(sum >target)
                       r--;
                   else
                       l++;
                   
                   }
           }//end while
        }//end for
        map<int,int>::iterator iter = diff.begin();
        return (*iter).second;
    }
};