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!
1). 双指针
1.) 双指针。时间复杂度:O(n),空间复杂度:O(n)
#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
vector<vector<int> > FindContinuousSequence(int target) {
//存放结果
vector<vector<int>> res;
//两个起点,相当于动态窗口的两边,根据其窗口内的值的和来确定窗口的位置和大小
int low = 1;
int high = 2;
int sum = 3;
while (high>low)
{
//由于是连续的,差为1的一个序列,那么求和公式是(a0+an)*n/2
sum = (low+high)*(high-low+1)/2;
//相等,那么就将窗口范围的所有数添加进结果集
if (sum==target)
{
vector<int> out;
for (int i=low;i<=high;i++)
{
out.push_back(i);
}
res.push_back(out);
low++;
}
//如果当前窗口内的值之和大于sum,那么左边窗口右移一下
if (sum > target)
{
low++;
}
//如果当前窗口内的值之和小于sum,那么右边窗口右移一下
if (sum < target)
{
high++;
}
}
return res;
}
};
int main()
{
int target = 100;
vector<vector<int> >res;
res = Solution().FindContinuousSequence(target);
system("pause");
return 0;
}
# -*- coding:utf-8 -*-
class Solution:
def FindContinuousSequence(self, target):
# write code here
res = []
low = 1
high = 2
Sum = 3
while high > low:
Sum = (high+low)*(high-low+1)//2
if Sum == target:
out = []
for i in range(low,high+1):
out.append(i)
res.append(out[:])
low+=1
if Sum > target:
low+=1
if Sum < target:
high+=1
return res
if __name__ == '_ main__':
target = 100
res = Solution().FindContinuousSequence(target)
print(res)