【LeetCode 0050】【分治/递归】求x的n次方

Jim-2ha02024-07-15 16:16

Pow(x, n)

Implement pow(x, n), which calculates x raised to the power n (i.e., x^n).

Example 1:

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**Input:** x = 2.00000, n = 10
**Output:** 1024.00000

Example 2:

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**Input:** x = 2.10000, n = 3
**Output:** 9.26100

Example 3:

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**Input:** x = 2.00000, n = -2
**Output:** 0.25000
**Explanation:** 2^-2 = 1/2^2 = 1/4 = 0.25

Constraints:

-100.0 < x < 100.0
-2^31 <= n <= 2^31-1
n is an integer.
Either x is not zero or n > 0.
-10^4 <= x^n <= 10^4

Idea

text 复制代码

* 调库函数Math.pow(x,n)
* 暴力乘法
* 分治递归，一分为二

JavaScript Solution

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/**
 * @param {number} x
 * @param {number} n
 * @return {number}
 */
var myPow = function(x, n) {
    if( !n ){
        return 1
    }

    if( n < 0 ){
        return 1/myPow(x,-n)
    }

    if( n%2 ){
        return x*myPow(x,n-1)
    }
    return myPow(x*x, n/2 ); 

};