Leetcode 16. Numerical integral power (power + fast recursive/iteration)

实现 pow(x, n) ,即计算 x 的 n 次幂函数（即,xⁿ）.不得使用库函数,同时不需要考虑大数问题.

示例 1：

输入：x = 2.00000, n = 10
输出：1024.00000
示例 2：

输入：x = 2.10000, n = 3
输出：9.26100
示例 3：

输入：x = 2.00000, n = -2
输出：0.25000
解释：2-2 = 1/22 = 1/4 = 0.25

提示：

-100.0 < x < 100.0
-2³¹ <= n <= 2³¹-1
-10⁴ <= xⁿ <= 10⁴

思路：

快速幂+递归

class Solution {
    
public:
    double myQuick(double x,long long k){
    
        if(k==0)
            return 1.0;
        double y=myQuick(x,k/2);
        return k%2==0 ? y*y : y*y*x;
    }
    double myPow(double x, int n) {
    
        long long N=n;
        return N>0 ? myQuick(x,N) : 1.0/myQuick(x,-N);
    }
};

快速幂+迭代：

class Solution {
    
public:
    double myPow(double x, int n) {
    
        if(n==0)
            return 1.0;
        if(x==0)
            return 0;
        double ans=1.0;
        long long b=n; //因为intWhen it is a positive number, the complement is itself,If it is negative number's complement,Need to add after changing the sign1,如果传进来的是（-2）的32次方,It will be exceeded after it is reversedint,所以要用long longto expand the type
        if(n<0){
    
            x=1/x; //Here is the direct reciprocal first,Calculate the countdown againn次方
            b=-b;
        }
        while(b>0){
    
            if(b&1==1)  //如果b最后一位为1,Just multiply in the final result,是0It doesn't count in the results
                ans*=x;
            x*=x;
            b>>=1;  //The right shift operator can be thought of as deleting the last bit,Remove the last one in the low order,高位正数补0,负数补1
        }
        return ans;
    }
};

You can see the details of the left shift and right shift operations：Algorithms for left shift and right shift