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 code:
Recursive:
class Solution {
public:
double myPow(double x, int n) {
if (n >= 0) return pow(x, n);
else return 1.0 / pow(x, n);
}
double pow(double x, int n) {
if (n == 0) {
return 1;
}
double y = pow(x, n / 2);
if (n % 2 == 0) {
return y * y;
} else {
return y * y * x;
}
}
};
Runtime: 4 ms, faster than 99.40% of C++ online submissions for Pow(x, n).
Iteration:
class Solution {
public:
double myPow(double x, int n) {
long y = abs((long)n);
double res = 1;
while (y > 0) {
if (y % 2 != 0) {
res *= x;
}
x *= x;
y /= 2;
}
if (n < 0) {
res = 1.0 / res;
}
return res;
}
};
Runtime: 4 ms, faster than 99.40% of C++ online submissions for Pow(x, n).