[Leetcode] [Javascript]3sum Closest

3Sum Closest 

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).

https://leetcode.com/problems/3sum-closest/

The same idea as the 3Sum http://www.cnblogs.com/Liok3187/p/4540921.html

Each round is fixed by a number, with a double pointer pointing to the head and tail of the array.

If the current three number is less than target, the head pointer + +, otherwise the tail pointer--until the pointer touches, records the closest result.

1 /**2 * @param {number[]} nums3 * @param {number} target4 * @return {number}5  */6 varThreesumclosest =function(nums, target) {7     varMin =Infinity;8     varres = 0;9Nums =nums.sort (sorting);Ten  One      for(vari = 0; i < nums.length-2; i++){ A         if(Nums[i-1]!== undefined && nums[i-1] = = =Nums[i]) { -             Continue; -         } the         varCurr =Nums[i]; -         varm = i + 1; -         varn = nums.length-1; -          while(M <N) { +             varTMP = Math.Abs (Nums[m] + nums[n] + Curr-target); -             if(TMP <min) { +Min =tmp; Ares = Nums[m] + nums[n] +Curr; at             } -  -             if(Nums[m] + nums[n] + Curr <target) { -m++; -}Else { -n--; in             } -         } to     } +     returnRes; -  the     functionsorting (A, b) { *         if(A >b) { $             return1;Panax Notoginseng}Else if(A <b) { -             return-1; the}Else{ +             return0; A         } the     } +};

[Leetcode] [Javascript]3sum Closest