leetcode: Two Sum

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#leetcode #堆排序

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本文介绍了一种解决两数之和问题的有效算法。通过排序和二分查找的方法，在整数数组中找到两个数使它们的和等于特定的目标值，并返回这两个数的下标。文章详细展示了使用堆排序进行预处理的过程，以及如何通过二分查找来提高搜索效率。

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

思路

先将输入的序列排序，然后对每一个数字，使用二分法寻找目标和与这个数字的差，如果找到则返回。用桶排序貌似更快，但是总感觉这个不是一个根本上的提升，其他的方法和思路目前还不知道。下面是accepted的代码，排序使用的是堆排序

class Solution{
public:
    vector<int> twoSum(vector<int> &numbers, int target){
        vector<int> res;
        res.push_back(0);
        res.push_back(0);
        if(numbers.size()<2) return res;
		vector<pair<int,int> > index_numbers;
		for(int i=0;i<numbers.size();i++)
		{
			index_numbers.push_back(make_pair(numbers[i],i));
		}
        heapSort(index_numbers);
        for(int i=0;i<index_numbers.size();i++)
        {
            int t=target-index_numbers[i].first;
            int ls=0,rs=index_numbers.size()-1;
            if(t>index_numbers[rs].first) continue;
            if(t<index_numbers[ls].first) continue;
            for(;ls<=rs;)
            {
                int ms=(ls+rs)/2;
                if(t<index_numbers[ms].first)
                    rs=ms-1;
                else if(t>index_numbers[ms].first)
                    ls=ms+1;
                else
                {
					if(ms==i)
					{
						if(ms>=1 && t==index_numbers[ms-1].first)
						{
							res[0]=index_numbers[ms-1].second+1;
							res[1]=index_numbers[i].second+1;
							if(res[0]>res[1]) swap(res[0],res[1]);
							return res;
						}
						if(ms<=index_numbers.size()-2 && t==index_numbers[ms+1].first)
						{
							res[0]=index_numbers[i].second+1;
							res[1]=index_numbers[ms+1].second+1;
							if(res[0]>res[1]) swap(res[0],res[1]);
							return res;
						}
						break;
					}
					res[0]=index_numbers[i].second+1;
					res[1]=index_numbers[ms].second+1;
					if(res[0]>res[1]) swap(res[0],res[1]);
                    return res;
                }
            }
        }
        return res;
    }

	inline void swap(int &l,int &r)
	{
		int tmp=r;r=l;l=tmp;
	}
    void heapBuild(vector<pair<int,int> > &numbers)
    {
        for(int i=numbers.size()-1;i>=0;i--)
        {
            heapAdjust(numbers,i,numbers.size()-i);
        }
    }

    void heapSort(vector<pair<int,int> > &numbers)
    {
        heapBuild(numbers);
        for(int i=numbers.size()-1;i>0;i--)
        {
            pair<int,int> tmp=numbers[0];
            numbers[0]=numbers[i];
            numbers[i]=tmp;
            heapAdjust(numbers,0,i);
        }
    }

    void heapAdjust(vector<pair<int,int> > &numbers, int cur,int length)
    {
        for(int i=cur;i<cur+length;)
        {
            int left=leftChild(i);
            int right=rightChild(i);
            if(left>=cur+length) return;
            int biggerChild=left;
            if(left<cur+length-1 && smallerThan(numbers[left],numbers[right])) biggerChild=right;
            if(biggerThan(numbers[biggerChild],numbers[i]))
            {
                pair<int,int> tmp=numbers[i];
                numbers[i]=numbers[biggerChild];
                numbers[biggerChild]=tmp;
            }
            i=biggerChild;
        }
    }

	bool smallerThan(pair<int,int> l,pair<int,int> r)
	{
		return l.first<r.first;
	}

	bool biggerThan(pair<int,int> l,pair<int,int> r)
	{
		return l.first>r.first;
	}

    inline int leftChild(int i) { return 2*i+1; }
    inline int rightChild(int i) { return 2*i+2; }
};