LeetCode 191. Number of 1 Bits
原题链接在这里：https://leetcode.com/problems/number-of-1-bits/
题目：
Write a function that takes an unsigned integer and return the number of '1' bits it has (also known as the Hamming weight).
Example 1:
Input: 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.
Example 2:
Input: 00000000000000000000000010000000 Output: 1 Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.
Example 3:
Input: 11111111111111111111111111111101 Output: 31 Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.
Note:
- Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer 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 above the input represents the signed integer
-3.
Follow up:
If this function is called many times, how would you optimize it?
题解：
Bit manipulation每次右移一位 & 1.
unsigned number 右移 >>>. signed number 右移>>.
也可以使用Integer.bitCount() function.
Time Complexity: O(1). Space: O(1).
AC Java:
1 public class Solution { 2 // you need to treat n as an unsigned value 3 public int hammingWeight(int n) { 4 //Method 1 5 // int count = 0; 6 // for(int i = 0; i<32; i++){ 7 // count += (n>>>i)&1; 8 // } 9 // return count; 10 11 //Method 2 12 int count = 0; 13 while(n != 0){ 14 count += (n&1); 15 n = n>>>1; 16 } 17 return count; 18 } 19 }
Method 3 是通过 n & (n-1)消掉最右侧的一个1.
消掉一个1, 对应的res就加一。
AC Java:
1 public class Solution { 2 // you need to treat n as an unsigned value 3 public int hammingWeight(int n) { 4 //Method 3 5 int res = 0; 6 while(n!=0){ 7 n = n & n-1; 8 res++; 9 } 10 return res; 11 } 12 }
