【LeetCode】461. Hamming Distance

    xiaoxiao2021-03-25  25

    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。

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