# Find power

### 1. Link

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### 2. 题目描述

给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方。\
\
保证base和exponent不同时为0。不得使用库函数，同时不需要考虑大数问题，也不用考虑小数点后面0的位数。\
\
示例1

### 输入

复制

```
2.00000,3
```

### 返回值

复制

```
8.00000
```

### 3. 思路

1. base^exponent = x^n = x^(  2^1 + 2^2 + .. + 2^n)= x^(2^1) \* x^(2^2)  \*x^(2^3)\* ...x^(2^n)
2. 而其中 x^(2^i) 就是累积的base， i 代表把exponent的二进制里面的第i位，如果第i位为1，就乘上这个项
3. 在while loop里面，每次计算 base \*=base 来计算 x^(2^i)
4. 之后可以通右移 exponent 的bit 看第一位bit是否为1，如果是就需要乘上对应的这个base
5. 考虑到 exponent 可以为负数，需要提前把exponent取反，最后把结果倒数
6. Time: O(logn), Space: O(1)&#x20;

   &#x20;

### 4. Coding

```
# -*- coding:utf-8 -*-
class Solution:
    def Power(self, base, exponent):
        #
        # write code here
        # base*base in iteration = base^1, base^2, base^4, base^8 
        #... = base^(2^0 + 2^1 + 2^2 + 2^3 + ....2^n)
        # res = x1* base^1 * x2*base^2 * x3*base^4* ...
        # xi = 1 or 0 bit depending on exponent 的bit 二进制位数
        #用于倒数
        sign = False
        if exponent <0:
            sign = True
            exponent *= -1
        res = 1
        # base^exp = base ^ (2^1 + 2^2 +.. 2^n)
        # = base^(2^1) * base^(2^2) * ... *base^(2^n)
        # 如果对于的base^(2^i) 是存在，就把那个base 乘上去 res里面
        while exponent >0:
            if exponent & 1 ==1:
                res = res * base
            exponent >>= 1
            # 计算每个base^(2^i)
            base *= base
        return 1/res if sign else res
```


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