3Sum Closest leetcode

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).
我的思路：
　　　　（1）首先对给定的数组进行排序，让里面的数值从小到大；
　　　　（2）遍历数组，用一个for循环控制作为外循环，下标从小到大进行遍历，因为需要3个数，所以用外循环的这个下标对应的数组值作为base，然后再
　　　　　　　　初始化一个begin作为base的下一位置，end作为数组末尾，三者相加得到sum，若是value=sum-target（目标值）更小则保存sum的值，并且若value<0
　　　　　　　　则begin++，若value>o则end--，若是value=0，则返回target；
看代码：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target); 
    int  findClosest(vector<int>& nums, int base, int end, int target); 
private:
    int result, finalResult, gap = 99999999, gap1 = 99999999;
};
int Solution::threeSumClosest(vector<int>& nums, int target){
    //int result,gap=99999999;
    sort(nums.begin(), nums.end());
    int len = nums.size();
    for (int i = 0; i < nums.size() - 2; i++)
    {
        //cout << "before" << endl;
        int res1 = findClosest(nums, i, len - 1, target);
        if (res1 < gap1)
            gap1 = res1;
        finalResult = result;
    }
    //    cout << "over" << endl;
    //    return gap1;
    return finalResult;

}
int Solution::findClosest(vector<int>& nums, int base, int end, int target){
    //cout << "executing" << endl;
    int begin = base + 1;
    while (begin < end){
        int value = nums[base] + nums[begin] + nums[end] - target;
        int temp = abs(nums[base] + nums[begin] + nums[end] - target);
        if (temp <= gap){
            //        int s = nums[base], s2 = nums[begin], s3 = nums[end];
            gap = abs(nums[base] + nums[begin] + nums[end] - target);
            result = nums[base] + nums[begin] + nums[end];
            if (value == 0)
                return 0;
            else if (value>0)
                end--;
            else if (value < 0)
                begin++;
        }
        else { 
            if (value>0) end--;
            else begin++;
        //    return gap; 
        }
    }
    return gap;
}