16.3Sum Closest (two-pointers)

Given an array S of n integers, find three integers in S such so the sum is closest to a give n number, target. Return the sum of the three integers. You may assume this each input would has exactly one solution.

    For example, given array S = {-1 2 1-4}, and target = 1.    The sum is closest to the target is 2. (-1 + 2 + 1 = 2).

classSolution { Public:    intThreesumclosest (vector<int>& Nums,inttarget) {        intSize =nums.size ();        Sort (Nums.begin (), Nums.end ()); Diff=Int_max;  for(inti =0; I < size-2; i++){            if(i>0&& nums[i]==nums[i-1])Continue; Find (Nums, I+1, size-1, target-Nums[i]); if(diff = =0)returnTarget; }        returntarget+diff; }          voidFind (vector<int>& Nums,intStartintEndinttarget) {        intsum;  while(start<end) {Sum= nums[start]+Nums[end]; if(Sum = =target) {diff=0; return; }            Else if(sum>target) {                 Do{End--; } while(End!=start && nums[end] = = nums[end+1]); if(Sum-target < ABS (diff)) diff = sum-Target; }            Else{                 Do{Start++; } while(start!= end && Nums[start] = = nums[start-1]); if(Target-sum < ABS (diff)) diff = Sum-target;//not only in the final check: there may be this situation, the previous sum>target, this sum<target, and the next time start==end, then it is likely that the previous sum is closer to the target than this sum            }        }    }Private:    intDiff//How much bigger is the sum};

16.3Sum Closest (two-pointers)