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]
AC代碼
class Solution {
public:
double myPow(double x, int n) {
if (x == 1 || x == 0) return x;
if (n == 0) return 1;
double ans = 1.0;
for (int i = n; i != 0; i /= 2) {
if (i % 2 != 0) ans *= x;
x *= x;
}
return n > 0 ? ans : 1 / ans;
}
};
總結(jié)
還有快速冪的相關(guān)解法禁舷，有時間再學(xué)
