https://leetcode.com/problems/powx-n/description/
Implement pow(x, n), which calculates x raised to the power n (xn).
Example 1:
Input: 2.00000, 10 Output: 1024.00000
Example 2:
Input: 2.10000, 3 Output: 9.26100
Example 3:
Input: 2.00000, -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25
Note:
- -100.0 < x < 100.0
- n is a 32-bit signed integer, within the range [−231, 231 − 1]
这题可是不能再错了啊…………
都写多少遍了
结果还是忘记了n<0
class Solution {
public:
double myPow(double x, int n) {
if(x==0.0)
return 0.0;
if(n==0)
return 1.0;
bool flag=1;
if(n<0)
flag=0;
double ans=1.0;
while(n){
if(n%2)
ans=ans*x;
n/=2;
x=x*x;
}
if(flag)
return ans;
else
return 1.0/ans;
}
};