461. Hamming Distance.md

461. Hamming Distance

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leetcode Daily Challenge on July 5th, 2020.

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Difficulty : Easy

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The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:

• 0 ≤ x, y < 231.

Example:

Input: x = 1, y = 4

Output: 2

Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑

The above arrows point to positions where the corresponding bits are different.

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Solution

• mine
  • Java
    • Runtime: 0 ms, faster than 100.00%, Memory Usage: 37.9 MB, less than 10.92% of Java online submissions

      // O(logN)time O(1)space
      // N = x ^ y;
      public int hammingDistance(int x, int y) {
          int res = 0;
          x = x ^ y;
          int count = 1;
          while(count < 32){
              res += (x & 1);
              x = x >> 1;
              count++;
          }
          return res;
      }
      

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• the most votes

  • Runtime: 0 ms, faster than 100.00%, Memory Usage: 33.2 MB, less than 5.09% of Java online submissions

    public int hammingDistance(int x, int y) {
         return Integer.bitCount(x ^ y);
    }
    

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