3Sum Closest--leetcode

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).

Test instructions is given an array, a target, in the array to find a ternary group and the nearest target value, the problem is the previous 3Sum deformation, the difference is that the problem is not repeating elements, and set the solution is unique.

My idea is: the same as the previous 3sum, the left and right one pointer, traversing the time, there is a back and right two cursors, left from the beginning of the i+1, from the array to start, if I, left, Right three elements plus and so on target, then can be added to the list of results, if sum-target>0, then the sum is larger, right should be moved to the left, and vice versa.

Talk is cheap>>

  　　  Public intThreesumclosest (int[] num,inttarget) {        if(num = =NULL|| Num.length < 3) {            return0; }        intMin =Integer.max_value; intRes=0;        Arrays.sort (num);  for(inti = 0; i < num.length-2; i++) {            intleft = i + 1; intright = Num.length-1;  while(Left <Right ) {                intsum = Num[i] + num[left] +Num[right];//System.out.printf (i+ "" +sum);                if(Sum = =target) {                    returnTarget; }                if(Min>=abs (sum-target)) {min= ABS (sum-target); Res=sum; }                if(sum-target>0) { Right--; }Else{ Left++; }            }        }        returnRes; }     Public intAbsinta) {        returnMath.Abs (a); }

3Sum Closest--leetcode