index.js

/*
 * @lc app=leetcode id=50 lang=javascript
 *
 * [50] Pow(x, n)
 *
 * https://leetcode.com/problems/powx-n/description/
 *
 * algorithms
 * Medium (27.42%)
 * Total Accepted:    303.1K
 * Total Submissions: 1.1M
 * Testcase Example:  '2.00000\n10'
 *
 * Implement pow(x, n), 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]
 * 
 * 
 */
/**
 * @param {number} x
 * @param {number} n
 * @return {number}
 */
var myPow = function(x, n) {
    if (n === 0) {
        return 1;
    }
    if (n < 0) {
        n = -n;
        x = 1 / x;
    }
    return (n % 2 === 0) ? myPow(x*x, n / 2) : x * myPow(x*x, Math.floor(n / 2));
};

console.log(myPow(2.00000, -2147483648), Math.pow(2.00000, -2147483648));

module.exports = {
    id:'50',
    title:'Pow(x, n)',
    url:'https://leetcode.com/problems/powx-n/description/',
    difficulty:'Medium',
}