JZ12 数值的整数次方
数值的整数次方
https://www.nowcoder.com/practice/1a834e5e3e1a4b7ba251417554e07c00?tpId=13&&tqId=11165&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking
2021年9月14日21:06:53
递归
public class Solution { public double Power(double base, int exponent) { if(exponent == 0) return 1; else if(exponent>0) return base * Power(base,exponent-1); else return 1/base*Power(base,exponent+1); } }
快速幂 k神
class Solution { public double myPow(double x, int n) { if(x == 0) return 0; long b = n; double res = 1.0; if(b < 0) { x = 1 / x; b = -b; } while(b > 0) { if((b & 1) == 1) res *= x; x *= x; b >>= 1; } return res; } }