O(1) Check Power of 2 - LintCode

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O(1) Check Power of 2 - LintCode

examination questions Using O(1) time to check whether an integer n is a power of 2. Example For n=4, return true; For n=5, return false; Challenge O(1) time 解题代码 class Solution { public: /* * @param n: An integer * @return: True or false */ bool che…

142. O(1) Check Power of 2【easy】

142. O(1) Check Power of 2[easy] Using O(1) time to check whether an integer n is a power of 2. Have you met this question in a real interview? Yes Example For n=4, return true; For n=5, return false; Challenge O(1) time 解法一: class Solution { public:…

Lintcode: O(1) Check Power of 2

Using O(1) time to check whether an integer n is a power of 2. Example For n=4, return true For n=5, return false Challenge O(1) time Tags Expand 这道题考察bit manipulation. 1的个数只能有1个才是power of 2. 主要是要注意Integer.MIN_VALUE,这个只有一个1,但是是false class Solution {…

142. O(1) Check Power of 2【LintCode by java】

Description Using O(1) time to check whether an integer n is a power of 2. Example For n=4, return true; For n=5, return false; Challenge O(1) time 题解:题目要求判断一个数是不是2的幂次方,并且规定时间复杂度为线性级.那么,就不能用常规的循环求解思路去解题了,之前做过一个不用加号求两数之和的题目.这两个题目有点类似,可以通过位运算来快速求解.先考虑这…

LintCode刷题笔记-- O(1) Check Power of 2

标签: 位运算题目: Using O(1) time to check whether an integer n is a power of 2. 解题思路: 这道题是利用位运算判断一个数是不是2 的整数次方思路:1. 如果一个数是2的整数次方的话,那么他的二进制表现形式上只有一位是1,其余的都会是02. 如果将n-1那么在原先所在的那一位就会变为0,其余之后的位数都会变成1,例如:16: 10000 ,15:011113. 将两者进行与运算,num&(num-1), 则所有的位上都会为0,这…

[LintCode]——目录

Yet Another Source Code for LintCode Current Status : 232AC / 289ALL in Language C++, Up to date (2016-02-10) For more problems and solutions, you can see my LintCode repository. I'll keep updating for full summary and better solutions. See cnblogs t…