实现 pow(x, n) ,即计算 x 的 n 次幂函数(即,xn)。
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
这是一道快速幂的题。
快速幂知识:https://liuyangjun.blog.csdn.net/article/details/85621386
class Solution {
public double myPow(double x, int n) {
long N=n;
return N>=0?quickMi(x,N):1.0/quickMi(x,-N);
}
public static double quickMi(double x,long N){
double res=1.0;
while(N>0){
if( N%2==1){
res=res*x;
}
x= x*x;
N>>=1;
}
return res;
}
}
力扣链接:https://leetcode-cn.com/leetbook/read/bit-manipulation-and-math/onqbav/



