50. Pow(x, n)中等 快速幂模板 两种写法
- Pow(x, n)
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
提示:
-100.0 < x < 100.0
-231 <= n <= 231-1
-104 <= xn <= 104
自递归,正负都可以计算
class Solution {
public double myPow(double x, int n) {
if(n==0)return 1;
double t=myPow(x,n/2);
if(n%2==1)return t*t*x;
if(n%2==-1)return t*t/x;
return t*t;
}
}
另写一个函数专门计算正幂
class Solution {
public double myPow(double x, int n) {
long N=n;//可能存在负数MAX导致溢出
if(N>0)return pow(x,N);
else return 1/pow(x,-N);
}
public double pow(double x,long n){
if(n==0)return 1;
double t=pow(x,n>>1);
if(n%2==0)return t*t;
return t*t*x;
}
}
迭代写法:
class Solution {
public double myPow(double x, int n) {
long N=n;//可能存在负数MAX导致溢出
if(N>0)return pow(x,N);
else return 1/pow(x,-N);
}
public double pow(double x,long n){
double t=x,ans=1.0;
while(n>0){
if(n%2==1)ans*=t;
t*=t;
n/=2;
}
return ans;
}
}
class Solution {
public double myPow(double x, int n) {
double ans=1.0,t=x;
while(n!=0){
if(n%2==1)ans*=t;
if(n%2==-1)ans/=t;
t*=t;
n/=2;
}
return ans;
}
}