【LeetCode】461. Hamming Distance

问题描述

问题链接：https://leetcode.com/problems/hamming-distance/#/description

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 < 2^31.

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. Subscribe to see which companies asked this question.

我的代码

探索过程

一开始想的是两个数异或，然后把异或出来的结果转成二进制的字符串（”01001”这样），然后数1的个数。后来觉得用位操作肯定更快啊，就准备这样写：

public class Solution { public int hammingDistance(int x, int y) { // 思路弄一个int当掩码， // 从0x01到0x80 // 每次左移一位去跟两个数与，如果结果不一样就加1 int mask = 0x01; int count = 0; while (mask <= 0x80){ if((x & mask) != (y & mask)){ count ++; } mask <<= 1; } return count; } }

我把一个16进制记成16位了，当然结果是没有通过。我仔细想了想，觉得知道问题在于位数了，所有改成了下面这样：

public class Solution { public int hammingDistance(int x, int y) { // 思路弄一个int当掩码， // 从0x00000001到0x80000000 // 每次左移一位去跟两个数与，如果结果不一样就加1 int mask = 0x00000001; int count = 0; while (mask <= 0x80000000){ if((x & mask) != (y & mask)){ count ++; } mask <<= 1; } return count; } }

这样了以后连1,4的测试用例都过不了了，我觉得很奇怪啊，凭什么啊。我怀疑是那个提交的系统出错了，就在IDE上试了试（嗯，没错，作弊了），确实是过不了。用IDE调试，发现0x80000000就是个负数。终于想起来符号位的事。改了过来。

通过的代码

public class Solution { public int hammingDistance(int x, int y) { // 思路弄一个int当掩码， // 从0x00000001到0x40000000(到这是因为再左移一位就成负数了)， // 每次左移一位去跟两个数与，如果结果不一样就加1 int mask = 0x00000001; int count = 0; while (mask <= 0x40000000 && mask > 0){ if((x & mask) != (y & mask)){ count ++; } mask <<= 1; } return count; } }

修改以后过了。可是运行速度只超过了5%的Java代码，再去评论区好好学习一个。

讨论区

来，见识一下大神们的风采。只能说大写的服！

Java 1 Line Solution :D

https://discuss.leetcode.com/topic/72093/java-1-line-solution-d

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

各种bitCount,来好好学习一个。https://tech.liuchao.me/2016/11/count-bits-of-integer/

the native way

int bitCount(int n) { int count = 0; for (int i = 0; i < 32; i++) { count += n & 1; n >>= 1; } return count; }

Brian Kernighan’s way

n & (n – 1) will clear the last significant bit set, e.g. n = 112

n | 1 1 1 0 0 0 0 n - 1 | 1 1 0 1 1 1 1 n &= n - 1 | 1 1 0 0 0 0 0 --------------------------- n | 1 1 0 0 0 0 0 n - 1 | 1 0 1 1 1 1 1 n &= n - 1 | 1 0 0 0 0 0 0 --------------------------- n | 1 0 0 0 0 0 0 n - 1 | 0 1 1 1 1 1 1 n &= n - 1 | 0 0 0 0 0 0 0

Hence loop will go through as many iterations as there are set bits, and when n becomes zero, count is exactly the answer.

int bitCount(int n) { int count = 0; while (n != 0) { n &= n - 1; count++; } return count; }

Java 3-Line Solution

https://discuss.leetcode.com/topic/72089/java-3-line-solution

public int hammingDistance(int x, int y) { int xor = x ^ y, count = 0; for (int i=0;i<32;i++) count += (xor >> i) & 1; return count; }

嗯，LeetCode值得好好刷刷。