求1+2+3+...+n,要求不能使用乘除法、for、while、if、else、switch、case等关键字及条件判断语句(A?B:C)。python
时间限制:1秒;空间限制:32768Kcode
利用逻辑运算的短路特性,做为递归的终止条件。递归
要注意python中逻辑运算符的用法:a and b,a为False返回a,a为True就返回b.it
Python代码:io
class Solution: def Sum_Solution(self, n): # write code here result = n a = n and self.Sum_Solution(n-1) result += a return result
C++代码:class
class Solution { public: int Sum_Solution(int n) { int ans = n; ans && (ans += Sum_Solution(n - 1)); return ans; } };