3 3Sum closest_Leetcode

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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).

To enumerate a subset of specified length without repetition, the best method is to keep the index increasing. (e.g. 0,1,2; 1,3,4)

The brute force enumeration use a 3-for loop which is O(n^3).

Remind the technique we used in 2Sum, we could sort first then start from the begin and the end to get the most closest sum.

So here we put the first num in a for loop, and use the technique in 2Sum to enumerate the other two num. The time complexity is O(n^2). Note the index of the elements in the subset is increasing. 

FIRST ERROR: Since we want to find the closest, I firstly wrongly used right-- to change the while loop index. How stupid error it is... I should change the index according to the sum compared with the target.

Code:

lass Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        int n = num.size();
        if(n < 3) return 0;
        int closest = INT_MAX;
        int mingap = INT_MAX;

        sort(num.begin(), num.end());
        for(int i = 0; i < n-2; i++)
        {
            int left = i+1, right = n-1;
            while(left < right)
            {
                int cursum = num[i] + num[left] + num[right];
                int gap = abs(target - cursum);
                if(gap < mingap)
                {
                    closest = cursum;
                    mingap = gap;
                }
                if(cursum < target) left++;   // first error
                else if(cursum > target) right--;
                else return target;
            }
        }
        return closest;
    }
};

3 3Sum closest_Leetcode

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原文地址：http://www.cnblogs.com/avril/p/4020698.html