题目来源:LeetCode: number-of-1-bits
问题描述:二进制串中1的个数,又被称为汉明重量
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.
方案1:
Note:从末尾开始,逐位统计1的个数;末位为1时为奇数,除2操作即为右移一位。
1 | int hammingWeight(uint32_t n) { |
方案2:
Note:从末尾开始,逐位统计1的个数,多用位操作的方式
1 | int hammingWeight(uint32_t n) { |
方案3:
Note:利用特性n&(n-1)的结果为n去除最右侧1后的数。循环次数相较方案1、2有效减少。
1 | int hammingWeight(uint32_t n) { |
方案4:
Note:维基百科中Hamming Weight的巧妙方法。
1 | int hammingWeight(uint32_t n) { |
这种方式暂时还没理解,可参考这篇文章进一步理解。