【Leetcode】3Sum Closest

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).


1st （6 tries ）

class Solution 
{
public:
    int threeSumClosest(vector<int> &num, int target) 
    {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int diff = INT_MAX;
        int sum;
        int ret;
        sort(num.begin(),num.end());
        for(int i = 0;i < num.size();++i)
        {
            int start = i + 1;
            int end = num.size() - 1;
            while(end > start)
            {
                sum = num[i] + num[start] + num[end];
                if(sum == target)
                    return sum;
                else
                {
                    int dis = abs(sum - target);
                    if(dis < diff)
                    {
                        diff = dis;
                        ret = sum;
                    }
                    if(sum > target)
                    {
                        end--;
                    }
                    if(sum < target)
                    {
                        start++;
                    }
                }
            }
        }
        return ret;
    }
};

　　

2nd （ 2 tries ）

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        sort(num.begin(),num.end());
        int first,second,third,mingap,ans;
        mingap = INT_MAX;
        for(first = 0;first < num.size() - 2;first++) {
            second = first + 1;
            third = num.size() - 1;
            while(second < third) {
                int sum = num[first] + num[second] + num[third];
                if(sum == target) {
                    return sum;
                }
                else if(sum > target) {
                    if(abs(sum-target) < mingap) {
                        mingap = abs(sum-target);
                        ans = sum;
                    }
                    third--;
                }
                else {
                    if(abs(sum-target) < mingap) {
                        mingap = abs(sum-target);
                        ans = sum;
                    }
                    second++;
                }
            }
        }
        return ans;
    }
};