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

 //第一种方法，找规律

 //第num开始，到num+n构成的序列和 == sum

 //么你会发现，则这些数全部减去(num-1)后构成序列 1,2,3,4,5,,,,n,而他们的和为n*(n+1)/2

 //以满足条件的序列必满足： n*(num-1) + n*(n+1)/2 == sum

 class Solution01 {

 public:

     vector<vector<int> > FindContinuousSequence(int sum) {

         vector<vector<int> >res;

         for (int i = ; i < sum; ++i)

         {

             int theta = ( * i - )*( * i - ) +  * sum;

             int gen = sqrt(theta);

             if (gen*gen != theta)continue;

             int a = ( -  * i) + gen;

             if (a %  != )continue;

             vector<int>temp;

             for (int j = ; j < a / ; ++j)

                 temp.push_back(i + j);

             res.push_back(temp);

         }

         return res;

     }

 };

 //第二种方法，滑动窗口

 //使用滑动窗口，和大了，则左指针右移，并吐出左数，和小了，则右指针右移，加上右数

 class Solution {

 public:

     vector<vector<int> > FindContinuousSequence(int sum) {

         if (sum < )return{};

         vector<vector<int> >res;

         int L = , R = , S =  + ;

         while (L < R && R < sum){

             if (S == sum){

                 vector<int>temp;

                 for (int j = L; j <= R; ++j)

                     temp.push_back(j);

                 res.push_back(temp);

             }

             if (S >= sum) {

                 S -= L;

                 ++L;

             }

             else {

                 ++R;

                 S += R;

             }

         }

         return res;

     }

 };