16.3Sum 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).

先对数组进行排序，对于数组中的每一个数i，设置左右指针j,k分别指向其后面的一个数和数组最后一个数，当j<k时，求数i,j,k三者之和，如果其和target相等，直接返回target。如果其和target差值更小，更新最终的返回值。如果sum比target更大， 则j向左移动一位，反之，则sum比target更小，则j向右移动一位。最终返回和target差值最小的那个sum。

1. class Solution {
   public:
       int threeSumClosest(vector<int>& nums, int target) {
           sort(nums.begin(),nums.end());
           int min=INT_MAX;
           int ret;
           for(int i=0;i<nums.size();i++){
              int j=i+1;
              int k=nums.size()-1;
              while(j<k){
                  int sum=nums[i]+nums[j]+nums[k];
                  int dif=abs(sum-target);
                  if(dif<min){
                      min=dif;
                      ret=sum;
                  }
                  if(dif==0)
                   return target;
                  else if(sum>target)
                   k--;
                  else
                   j++;
              }
           }
           return ret;
       }
   };