1. 程式人生 > >leetcode 50. Pow(x, n)【快速冪】

leetcode 50. Pow(x, n)【快速冪】

Implement pow(xn), 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]

這題可是不能再錯了啊…………

都寫多少遍了

結果還是忘記了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;
    }
};