leetcode笔记：Power of two

一. 题目描述

Given an integer, write a function to determine if it is a power of two.

二.题目分析

该题要求简单，给定一个整数，判断其是不是2的整数次幂，这道题的解题关键是找到一个规律：如果一个数字是2的整数次幂，若将该数写为二进制数，这个二进制数中有且仅有一位为1，其余均为0。根据这一性质，不难给出以下给出两种解决方法。

三.示例代码

// 将n不停右移，比较二进制数中1出现的个数，count = 1时判定为ture
bool isPowerOfTwo(int n) {
    int count = 0;
    while (n > 0)
    {
        count+=(n&0x01);
        n>>=1;
    }
    if(sum==1)
        return true;
    else
        return false;
}

以下方法同样利用了一个2的整数次幂的二进制写法中有且仅有一位为1的性质。假设该数为n，根据这一性质，则(n - 1)必然是将n的最高位1置0，然后其余二进制位均置1，当且仅当这种情况下，有(n&(n-1)) == 0，一个例子：

n = 8 -> 1000

n - 1 = 7 -> 0111

则有：1000 & 0111 = 0000

class Solution {
public:
    bool isPowerOfTwo(int n) {
        return (n>0) && (!(n&(n-1)));
    }
};

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时间： 2024-08-26 04:41:50

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