# [LeetCode] 50. Pow(x, n) 求x的n次方

Implement pow(xn), which calculates x raised to the power n(xn).

Example 1:

Input: 2.00000, 10
Output: 1024.00000


Example 2:

Input: 2.10000, 3
Output: 9.26100


Example 3:

Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25


Note:

• -100.0 < x < 100.0
• n is a 32-bit signed integer, within the range [−231, 231 − 1]

class Solution {
public:
double myPow(double x, int n) {
if (n == 0) return 1;
double half = myPow(x, n / 2);
if (n % 2 == 0) return half * half;
if (n > 0) return half * half * x;
return half * half / x;
}
};

class Solution {
public:
double myPow(double x, int n) {
double res = 1.0;
for (int i = n; i != 0; i /= 2) {
if (i % 2 != 0) res *= x;
x *= x;
}
return n < 0 ? 1 / res : res;
}
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/50

Sqrt(x)

Super Pow

https://leetcode.com/problems/powx-n/

https://leetcode.com/problems/powx-n/discuss/19733/simple-iterative-lg-n-solution

https://leetcode.com/problems/powx-n/discuss/19546/Short-and-easy-to-understand-solution

https://leetcode.com/problems/powx-n/discuss/19544/5-different-choices-when-talk-with-interviewers

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posted @ 2015-04-01 14:34  Grandyang  阅读(33274)  评论(19编辑  收藏  举报