16.3Sum Closest (Two-Pointers)

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).

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int size = nums.size();

        sort(nums.begin(), nums.end());
        diff = INT_MAX;
        for(int i = ; i < size-; i++){
            if(i> && nums[i]==nums[i-])  continue;
            find(nums, i+, size-, target-nums[i]);
            if(diff == ) return target;
        }
        return target+diff;
    }

      void find(vector<int>& nums, int start, int end, int target){
        int sum;
        while(start<end){
            sum = nums[start]+nums[end];
            if(sum == target){
                diff = ;
                return;
            }
            else if(sum>target){
                do{
                    end--;
                }while(end!=start && nums[end] == nums[end+]);
                if(sum-target < abs(diff)) diff = sum - target;
            }
            else{
                do{
                    start++;
                }while(start!= end && nums[start] == nums[start-]);
                if(target - sum < abs(diff)) diff = sum - target; //不能只在最后检查：可能会有这种情况，前一次sum>target,这次sum<target，而且下次就start==end，那么很可能前一次的sum比这次的sum更接近target
            }
        }
    }
private:
    int diff; //how much bigger is the sum
};