Implement pow(x, n). 题意:计算x的次方 思路:这题的思路和sqrt的类似,向二分靠近.例如求4^5,我们可以这样求:res=4.4*4^4.就是将每次在res的基础上乘以x本身,换成,每次以x*=x的方式前行,这样时间复杂度为O(logn),代码如下: class Solution { public: double pow(double x, int n) { double res=1.0; ;i/=) { !=) res*=x; x*=x; } ?/res:res; }…

Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array. Example1: a = 2 b = [3] Result: 8 Example2: a = 2 b = [1,0] Result: 1024 Credits:Special thanks to @Stomac…

Given an integer (signed 32 bits), write a function to check whether it is a power of 4. Example: Given num = 16, return true. Given num = 5, return false. Follow up: Could you solve it without loops/recursion? Credits:Special thanks to @yukuairoy fo…

Given an integer, write a function to determine if it is a power of three. Follow up:Could you do it without using any loop / recursion? Credits:Special thanks to @dietpepsi for adding this problem and creating all test cases. 这道题让我们判断一个数是不是3的次方数,在Le…

Given an integer, write a function to determine if it is a power of two. Hint: Could you solve it in O(1) time and using O(1) space? 这道题让我们判断一个数是否为2的次方数,而且要求时间和空间复杂度都为常数,那么对于这种玩数字的题,我们应该首先考虑位操作 Bit Operation.在LeetCode中,位操作的题有很多,比如比如Repeated DNA Seque…

Implement pow(x, n). 这道题让我们求x的n次方,如果我们只是简单的用个for循环让x乘以自己n次的话,未免也把LeetCode上的想的太简单了,一句话形容图样图森破啊.OJ因超时无法通过,所以我们需要优化我们的算法,使其在更有效的算出结果来.我们可以用递归来折半计算,每次把n缩小一半,这样n最终会缩小到0,任何数的0次方都为1,这时候我们再往回乘,如果此时n是偶数,直接把上次递归得到的值算个平方返回即可,如果是奇数,则还需要乘上个x的值.还有一点需要引起我们的注意的是n有可能…

Given an integer (signed 32 bits), write a function to check whether it is a power of 4. Example:Given num = 16, return true. Given num = 5, return false. Follow up: Could you solve it without loops/recursion? 题目标签:Bit Manipulation 这道题目让我们判断一个数字是不是4的…

版权声明: 本文为博主Bravo Yeung(知乎UserName同名)的原创文章,欲转载请先私信获博主允许,转载时请附上网址 http://blog.csdn.net/lzuacm. Leetcode 372. 超级次方 - 题解 372.Super Pow 在线提交: https://leetcode.com/problems/super-pow/ 题目描述 你的任务是计算 ab" role="presentation">abab 对 1337 取模,a 是一个正整…

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:…

Given an integer, write a function to determine if it is a power of two. Example 1: Input: 1 Output: true Example 2: Input: 16 Output: true Example 3: Input: 218 Output: false 给一个整数,写一个函数来判断它是否为2的次方数. 利用计算机用的是二进制的特点,用位操作,此题变得很简单. 2的n次方的特点是:二进制表示中最高位是…