思路:
// @Title: Pow(x, n) (Pow(x, n))
// @Author: qisiii
// @Date: 2022-02-20 14:51:34
// @Runtime: 0 ms
// @Memory: 40.3 MB
// @comment:
// @flag:
class Solution {
public double myPow(double x, int n) {
if(n == 0) return 1;
if(n == 1) return x;
if(n == -1) return 1 / x;
double half = myPow(x, n / 2);
double mod = myPow(x, n % 2);
return half * half * mod;
}
}
+++ title = “Pow(x, n) (Pow(x, n))” draft = false +++
思路:模除作为结果
// @Title: Pow(x, n) (Pow(x, n))
// @Author: qisiii
// @Date: 2022-02-20 14:56:37
// @Runtime: 0 ms
// @Memory: 40.6 MB
// @comment: 模除作为结果
// @flag: BLUE
class Solution {
public static double myPow(double x, int n) {
if (n == 0) {
return 1;
}
if (n == -1) {
return 1 / x;
}
if (n == 1) {
return x;
}
double v = myPow(x, n / 2);
//模除不只有1和0,还有-1
double mod = myPow(x, n % 2);
return v*v*mod;
}
}