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Pow(x, n) #82

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louzhedong opened this issue Nov 14, 2018 · 0 comments
Open

Pow(x, n) #82

louzhedong opened this issue Nov 14, 2018 · 0 comments

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@louzhedong
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习题

出处:LeetCode 算法第50题

实现 pow(x, n) ,即计算 x 的 n 次幂函数。

示例 1:

输入: 2.00000, 10
输出: 1024.00000

示例 2:

输入: 2.10000, 3
输出: 9.26100

示例 3:

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] 。

思路

1.直接使用Math.pow

2.采用递归的方式

解答

/**
 * @param {number} x
 * @param {number} n
 * @return {number}
 */
var myPow = function (x, n) {
  if (n == 0) return 1;
  if (n == 1) return x;
  if (n == -1) return 1 / x;
  if (n % 2 == 0) {
    var temp = myPow(x, n / 2);
    return temp * temp;
  } else {
    return x * myPow(x, n - 1);
  }
};

console.log(myPow(2.1,3));
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