Leetcode--Number of 1 Bits
2015-03-16 20:14
330 查看
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.
思路:首先想到的是将比特位逐个右移,然后计算1的数量
具体代码如下:int hammingWeight(uint32_t n) { int res = 0; while (n != 0){//当数字n不为0,或者是说当n中还有1的时候,一直循环 if(n %2 != 0)//判断n是否能够被2整除就是判断右移之后的n的末位是不是1,如果是1的话,计数器加1 ++res; n/=2;//n右移一位 } return res; }还有一种相似的做法就是逐个对n的32比特位进行与的运算,然后算出1的数量
代码如下:
int hammingWeight(uint32_t n) { int res = 0; for (int i = 0; i < 32; ++i) { res += (n & 1);//n&1 的意思就是判断n的末位是不是1,是1的话,结果为1,不是的话结果为0,然后再累加到res中 n = n >> 1;//n右移一位 } return res; }但是这还是不够快,要循环32遍才能找出所有的1,如果能有多少个1循环多少遍,就是最快的方法。
举例来说:
n = 0x110100 n-1 = 0x110011 n&(n - 1) = 0x110000
n = 0x110000 n-1 = 0x101111 n&(n - 1) = 0x100000
n = 0x100000 n-1 = 0x011111 n&(n - 1) = 0x0
由上可知,n&(n-1)能够把除了n和n-1中共有的1保留,其他的全部变0.由此写出代码:
<pre name="code" class="cpp">int hammingWeight(uint32_t n) { int re = 0; int left = 0; <span style="white-space:pre"> </span>while(0 != n) <span style="white-space:pre"> </span>{ <span style="white-space:pre"> </span> n = n&(n - 1); <span style="white-space:pre"> </span> ++re; <span style="white-space:pre"> </span>} }
这样便可做到有多少1循环多少遍。
参考文章:
点击打开链接
点击打开链接
相关文章推荐
- Leetcode: Number of 1 Bits
- leetcode- Number of 1 bits, power of two, power of three
- [Leetcode]-Number of 1 Bits
- LeetCode(191) Number of 1 Bits
- [LeetCode]Number of 1 bits
- LeetCode191 Number of 1 Bits
- Leetcode 191 Number of 1 Bits
- LeetCode 191 number of 1 bits
- leetcode --Number of 1 Bits
- LeetCode 191:Number of 1 Bits
- [LeetCode] Number of 1 Bits
- LeetCode 191: Number of 1 Bits
- [leetcode][javascript]Number of 1 Bits
- [LeetCode][191][Number of 1 Bits]
- LeetCode 191 Number of 1 Bits(1 比特的数字们)
- [LeetCode] Number of 1 Bits
- leetcode:191 Number of 1 Bits-每日编程第十三题
- Leetcode Number of 1 Bits
- Leetcode Problem.191—Number of 1 Bits
- LeetCode191:Number of 1 Bits