剑指 Offer 16. 数值的整数次方
本问题对应的 leetcode 原文链接:剑指 Offer 16. 数值的整数次方
问题描述
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
解题思路
视频讲解直达: 本题视频讲解
代码实现
class Solution {
public double myPow(double x, int n) {
double res = 1;
long y = n;
if(n < 0){
y = -y;
x = 1 / x;
}
while(y > 0){
if(y % 2 == 1){
res = res * x;
}
x = x * x;
y = y / 2;
}
return res;
}
}
时间复杂度:O(logn)
空间复杂度:O(1)
Python
class Solution(object):
def myPow(self, x, n):
"""
:type x: float
:type n: int
:rtype: float
"""
res = 1
y = n
if n < 0:
y = -y
x = 1 / x
while y > 0:
if y % 2 == 1:
res = res * x
x = x * x
y = y // 2
return res
C++
class Solution {
public:
double myPow(double x, int n) {
double res = 1;
long y = n;
if (n < 0) {
y = -y;
x = 1 / x;
}
while (y > 0) {
if (y % 2 == 1) {
res = res * x;
}
x = x * x;
y = y / 2;
}
return res;
}
};
Go
func myPow(x float64, n int) float64 {
res := 1.0
y := int64(n)
if n < 0 {
y = -y
x = 1 / x
}
for y > 0 {
if y % 2 == 1 {
res *= x
}
x *= x
y /= 2
}
return res
}
JS
var myPow = function(x, n) {
let res = 1;
let y = n;
if (n < 0) {
y = -y;
x = 1 / x;
}
while (y > 0) {
if (y % 2 === 1) {
res = res * x;
}
x = x * x;
y = Math.floor(y / 2);
}
return res;
};