50. Pow(x, n)

Implement pow(x, n).

Solution1：递归实现

思路： 2^7 = (half=2^3) ^2 * 2，分成相同的两半2^{3，只计算其中一半即可，而不用2来乘七次，继续对2}3递归相同操作
实现用递归
Time Complexity: O(logN) Space Complexity: O(logN) 递归缓存

Solution2：Iterative实现

思路： x ^ 5, 5=101, 低到高第一位是1, result *= x, 第二位是0， nothing for x^2，第三位是1，result *= x^4，而x ^ n是累积乘自身得出的 x^4 = (x^2) ^ 2
实现用递归
Time Complexity: O(logN) Space Complexity: O(1)

Solution1 Code：

class Solution1a {
    // 2^7 = (half=2^3)^2 * 2
    // 2^6 = (half=2^3)^2
    public double myPow(double x, int n) {
        if(n < 0) {
            n = -n;
            x = 1 / x;
        }
        return helper(x, n);
    }
    private double helper(double x, int n) {
        if(n == 0) return 1;
        double half = helper(x, n / 2);
        if(n % 2 == 0)
            return half * half;
        else 
            return x * half * half;
          
    }
}

public class Solution1b{
    public double pow(double x, int n) {
        if(n == 0)
            return 1;
        if(n<0){
            n = -n;
            x = 1/x;
        }
        return (n%2 == 0) ? pow(x*x, n/2) : x*pow(x*x, n/2);
    }
}

Solution2 Code：

class Solution2 {
    double myPow(double x, int n) { 
        if(n == 0) return 1;
        if(n < 0) {
            n = -n;
            x = 1/x;
        }
        double ans = 1;
        while(n > 0){
            if(n & 1) ans *= x;
            x *= x;
            n >>= 1;
        }
        return ans;
    }
}