剑指offer ：和为S的连续正数序列

题目：小明很喜欢数学,有一天他在做数学作业时,要求计算出9~16的和,他马上就写出了正确答案是100。但是他并不满足于此,他在想究竟有多少种连续的正数序列的和为100(至少包括两个数)。没多久,他就得到另一组连续正数和为100的序列:18,19,20,21,22。现在把问题交给你,你能不能也很快的找出所有和为S的连续正数序列? Good Luck!
输出描述:
输出所有和为S的连续正数序列。序列内按照从小至大的顺序，序列间按照开始数字从小到大的顺序

//a1 = [i,(sum/2) + 1]，n = [1, (sum/2) + 2 - a1)  计算(a1 + an)*n/2 是否等于sum,等于则返回一组结果
class Solution {
public:
    vector<vector<int> > FindContinuousSequence(int sum) 
    {
        vector<vector<int> > res;
        for (int i=1; i<(sum/2) + 1; i++)
        {
            int n=0;
            if (find_n(sum, i, n))
            {
                vector<int> tmp;
                for (int j=0;j<=n;j++)
                    tmp.push_back(i+j);
                res.push_back(tmp);
            }
        }
        return res;
    }
    
    bool find_n(int sum, int start, int &n)
    {
        for (int i=1; i<=(sum/2) + 1 - start; i++)
        {
            if ((start + start + i)*(i + 1)/2 == sum)
            {
                n=i;
                return true;
            }
        }
        return false;
    }
};

//二分法优化find_n
class Solution {
public:
    vector<vector<int> > FindContinuousSequence(int sum) 
    {
        vector<vector<int> > res;
        for (int i=1; i<(sum/2) + 1; i++)
        {
            int n=0;
            if (find_n(sum, i, n))
            {
                vector<int> tmp;
                for (int j=0;j<=n;j++)
                    tmp.push_back(i+j);
                res.push_back(tmp);
            }
        }
        return res;
    }
    
    bool find_n(int sum, int start, int &n)
    {
        int left = 1, right = (sum/2) + 1 - start, middle = 0, tmp = 0;
        while(left <= right)
        {
            middle = (left + right)/2;
            tmp = (start + start + middle)*(middle + 1)/2;
            if (tmp == sum)
            {
                n = middle;
                return true;
            }
            else if (tmp < sum)
            {
                left = middle + 1;
            }
            else
            {
                right = middle - 1;
            }
        }
        return false;
    }
};


/*维护left  和 big  cur_sum 类似于两个数之和；当前和小于target，big++,cur_sum+=big, 当前和大于target，cur_sum -= left，left++, 当前和等于target，left-big是一个结果，cur_sum -= left，left++*/
class Solution {
public:
    vector<vector<int> > FindContinuousSequence(int sum) 
    {
        vector<vector<int> > res;
        if (sum<=2)
            return res;
        int left=1, right=2, cur_sum=3, left_max=sum/2+1;
        while(left < left_max)
        {
            if (cur_sum < sum)
            {
                cur_sum += ++right;
            }
            else if(cur_sum > sum)
            {
                cur_sum -= left++;
            }
            else
            {
                vector<int> tmp;
                for(int i=left;i<=right;i++)
                    tmp.push_back(i);
                res.push_back(tmp);
                cur_sum -= left++;
            }
        }
        return res;
    }
};