实现 pow(x, n) ,即计算 x 的 n 次幂函数。
示例
输入: 2.00000, 10
输出: 1024.00000
示例
输入: 2.00000, -2
输出: 0.25000
解释: 2-2 = 1/22 = 1/4 = 0.25
说明:
-100.0 < x < 100.0
n 是 32 位有符号整数,其数值范围是 −231, 231 − 1 。
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/powx-n 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
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class Solution {
public:
double myPow(double x, int n) {
if(n == 0) return 1.0;
long N = n;
if(N < 0)
{
x = 1/x;
N = -N;
}
return POW(x, N);
}
double POW(double x, int N)
{
if(N == 0) return 1.0;
double half = POW(x, N/2);
if(N%2 == 0)
return half*half;
else
return half*half*x;
}
};
class Solution {//超时,0.00001,2147483647
unordered_map<int ,double> m;
public:
double myPow(double x, int n) {
if(n == 0)
return 1.0;
long N = n;
if(n < 0)
{
x = 1/x;
N = -(long)n;//INT_MIN换号后溢出
}
return MyP(x, N);
}
double MyP(double x, long N)
{
if(N == 0)
return 1.0;
if(N%2 == 0)
{
if(m.count(N))
return m[N];
double a = MyP(x, N/2)*MyP(x, N/2);
m[N] = a;
return a;
}
else
{
if(m.count(N))
return m[N];
double b = MyP(x, N/2)*MyP(x, N/2)*x;
m[N] = b;
return b;
}
}
};
class Solution {
public:
double myPow(double x, int n) {
if(n == 0) return 1.0;
long N = n;
if(N < 0)
{
x = 1/x;
N = -N;
}
double ans=1;
double product = x;
for(long i = N; i; i /= 2)
{
if(i % 2 == 1)
ans *= product;
product *= product;
}
return ans;
}
};