【LeetCode】461. Hamming Distance

题目描述

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.

解决方案1 - C

int hammingDistance(int x, int y) {
    
    int tmpInt=x^y;
    int dis=0;
    
    while(tmpInt)
    {
        if((tmpInt>>1)<<1 != tmpInt)
        {
            ++dis;
        }
        
        tmpInt>>=1;
    }
    
    return dis;
} 

解题思路
异或：相同为0，相异为1.
利用异或的特性，先将两数x和y做异或运算。得到的结果tmpInt中，为1的位表示x和y中不相同的位置。将tmpInt先右移再左移，此时若tmpInt还等于原来的tmpInt，则表示最低位为0，否则为1。记录tmpInt中1的个数，即x和y不同位的个数。

解决方案2

int hammingDistance(int x1, int x2) {  
    int b1[32];
    int b2[32];  
    int r1 = x1;
    int r2 = x2;
    int i = 0, j = 0;
    while(r1 > 0){
        b1[i++] = r1 % 2;
        r1 = r1 >> 1;
    }
    while(r2 > 0){
        b2[j++] = r2 % 2;
        r2 = r2 >> 1;
    }
    int ans = 0;
    for(int k = 0; k < i && k < j; k++){
        if(b1[k] != b2[k]){
            ans++;
        }
    }
    if(i > j){
        for(int k = j; k < i; k++){
            if(b1[k] != 0){
                ans++;
            }
        }
    }else{
        for(int k = i; k < j; k++){
            if(b2[k] != 0){
                ans++;
            }
        }
    }  
    return ans;   
}

解题思路
将十进制的x和y的二进制形式存放到b数组中(每次除以2的余数)，然后比较两个b数组中不同位数的个数，最后注意不同长度的问题(位数较多的数，记录长出的部分1的个数)。

解决方案 - java

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

解题思路

What does come to your mind first when you see this sentence "corresponding bits are different"? Yes, XOR! Also, do not forget there is a decent function Java provided: Integer.bitCount().

bitCount实现几种实现方式(牛逼)

[Algorithm 101] Bit Manipulation: Count Bits of Integer (Hamming weight)

Java中几种位移运算符<< >> >>>