191. Number of 1 Bits
191. Number of 1 Bits
题目
Write a function that takes the binary representation of an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).
Note:
- Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
- In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 3 , the input represents the signed integer.
-3.
Example 1:
Input: n = 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.
Example 2:
Input: n = 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.
Example 3:
Input: n = 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.
Constraints:
- The input must be a binary string of length
32.
Follow up: If this function is called many times, how would you optimize it?
题目大意
编写一个函数，输入是一个无符号整数（以二进制串的形式），返回其二进制表达式中数字位数为 '1' 的个数（也被称为 汉明重量）。
解题思路
思路一：循环
可以直接循环检查给定整数 n 的二进制位的每一位是否为 1。
当检查第 i 位时，可以让 n 与 2^i 进行 与 运算，当且仅当 n 的第 i 位为 1 时，运算结果不为 0。
时间复杂度：O(k)，其中 k=32，一共需要检查 32 位。
空间复杂度：O(1)，只需要常数的空间保存若干变量。
思路二：位运算
由于 n & (n−1) 会 n 的二进制位中的最低位的 1 变为 0 ，如：6 & (6−1) = 4, 6 = (110)_2, 4 = (100)_2，运算结果 4 即为把 6 的二进制位中的最低位的 1 变为 0 之后的结果。
可以利用这个位运算的性质，不断让当前的 n 与 n - 1 做与运算，直到 n 变为 0 即可。因为每次运算会使得 n 的最低位的 1 被翻转，因此运算次数就等于 n 的二进制位中 1 的个数。
代码
/**
* @param {number} n - a positive integer
* @return {number}
*/
var hammingWeight = function (n) {
let res = 0;
for (let i = 0; i < 32; i++) {
if (n & (1 << i)) {
res++;
}
}
return res;
};
/**
* @param {number} n - a positive integer
* @return {number}
*/
var hammingWeight = function (n) {
let res = 0;
while (n) {
n &= n - 1;
res++;
}
return res;
};
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