LeetCode191 Number of 1 Bits. LeetCode231 Power of Two. LeetCode342 Power of Four
2016-12-12 20:05
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位运算相关 三道题
231. Power of Two
Given an integer, write a function to determine if it is a power of two. (Easy)分析:
数字相关题有的可以考虑用位运算,例如&可以作为筛选器。
比如n & (n - 1) 可以将n的最低位1删除,所以判断n是否为2的幂,即判断(n & (n - 1) == 0)
代码:
1 class Solution { 2 public: 3 bool isPowerOfTwo(int n) { 4 if (n <= 0) { 5 return false; 6 } 7 return ( (n & (n - 1)) == 0); //按位筛选,考虑位操作 8 } 9 };
191. Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight).For example, the 32-bit integer ’11' has binary representation
00000000000000000000000000001011, so the function should return 3. (Easy)
分析:
同上一个题,n & n-1可以去掉n的最低位1,因此循环即可求出1的个数。
代码:
1 class Solution { 2 public: 3 int hammingWeight(uint32_t n) { 4 int num = 0; 5 while (n > 0) { 6 n = (n & (n - 1)); 7 num++; 8 } 9 return num; 10 } 11 };
342. Power of Four
Given an integer (signed 32 bits), write a function to check whether it is a power of 4. (Easy)分析:
首先是4的幂肯定是2的幂,所以可以利用Power of Two, 其次考察 4, 16等数,其1出现在奇数位。
所以利用0101 &该数,可以删选掉是2的幂但不是4的幂。
代码:
1 class Solution { 2 public: 3 bool isPowerOfFour(int num) { 4 if (num <= 0) { 5 return false; 6 } 7 return ( (num & (num - 1)) == 0 && (num & 0x55555555) ); //有一个1,且1在奇数位上,&操作起到筛选器作用,5即0101 8 } 9 };
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