Leetcode's 16_3sum closest

Original title:

Given an array S of n integers, find three integers in s such so the sum is closest to a Given number, target. Return the sum of the three integers. You may assume this each input would have 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).

The analysis:

And the 3sum similar to the question number 15, but this problem requires three numbers and is a given number, that is, find a+b+c=target, solve the same ideas, the two number of the sum of the target into target-a can.

Problem-Solving Code:

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). Finds three digits from the given array, which is closest to the given target #include <iostream> #include <vector> #include <algorithm> using


namespace Std; Class Solution {public:int threesumclosest (vector<int>& nums, int target) {if (Nums.size () <3) {RE
		turn-1;
		int tarsum = 0;
		int tempsum = 0;
		int nmin = 1024;

		int nret = 0;  Sort (Nums.begin (), Nums.end ());  Causes the number in the container to be an ordered sequence for (int i=0; I<nums.size ()-2; i++) {tarsum = target-nums[i];//Declare the target of the search and A+b+c = targets, a+b =  target-c int j = i+1;

			Declares the search for the end-node int k=nums.size ()-1;

				while (j<k) {tempsum = Nums[j]+nums[k]-tarsum;
					if (ABS (TempSum) < nmin) {nmin = ABS (TempSum);
				Nret = Nums[i]+nums[j]+nums[k];
				} if (TempSum < 0) {j + +;
				else if (TempSum > 0) {k--; else if (tempsum = 0) {return TARget;
	}} return nret;
 }
};