使用for循环来获得2个字符串之间的汉明距离 [英] Using for loop to get the Hamming distance between 2 strings

问题描述

在这个任务中，我需要得到两个字符串sequence1和sequence2之间的汉明距离（两个相等长度的字符串之间的汉明距离是相应符号不同的位置数 - 来自维基百科）。

首先，我制作了2个新的字符串，它们是2个原始字符串，但是两个字母都是降低的，以便比较容易。然后我诉诸使用for循环，如果比较2个字符串。对于这两对字符串中的任何字符差异，循环会将1加到int x = 0。方法的返回值将是这个x的值。

  public static int getHammingDistance（String sequence1，String sequence2）{
 int a = 0; 
 String sequenceX = sequence1.toLowerCase（）; 
 String sequenceY = sequence2.toLowerCase（）; （int x = 0; x  for（int y = 0; y  if（sequenceX.charAt（x）== sequenceY.charAt（y））{
a + = 0; 
} else if（sequenceX.charAt（x）！= sequenceY.charAt（y））{
 a + = 1; 
} 
} 
} 
 return a; 
} 
  

所以代码看起来不错，任何我可以修复或优化代码？提前致谢。我是一个巨大的noob，所以请原谅我，如果我问什么愚蠢的解决方案

从我的观点下面的实现将是好的： / p>

  public static int getHammingDistance（String sequence1，String sequence2）{
 char [] s1 = sequence1.toCharArray（）; 
 char [] s2 = sequence2.toCharArray（）; 
 
 int short = Math.min（s1.length，s2.length）; 
 int longest = Math.max（s1.length，s2.length）; 
 
 int result = 0; 
 for（int i = 0; i  if（s1 [i]！= s2 [i]）result ++; 
} 
 
结果+ =最长 - 较短; 
 
返回结果; 
 
 
 
 
 $ b 
使用数组，避免调用两个方法（charAt）用于每个需要比较的单个字符; 
 
 
当一个字符串长于另一个字符串时，避免异常
 
 
 
In this task i need to get the Hamming distance (the Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different - from Wikipedia) between the two strings sequence1 and sequence2. 

First i made 2 new strings which is the 2 original strings but both with lowered case to make comparing easier. Then i resorted to using the for loop and if to compare the 2 strings. For any differences in characters in these 2 pair of string, the loop would add 1 to an int x = 0. The returns of the method will be the value of this x. 
public static int getHammingDistance(String sequence1, String sequence2) {
    int a = 0;
    String sequenceX = sequence1.toLowerCase();
    String sequenceY = sequence2.toLowerCase();
    for (int x = 0; x < sequenceX.length(); x++) {
        for (int y = 0; y < sequenceY.length(); y++) {
            if (sequenceX.charAt(x) == sequenceY.charAt(y)) {
                a += 0;
            } else if (sequenceX.charAt(x) != sequenceY.charAt(y)) {
                a += 1;
            }
        }
    }
    return a;
}
So does the code looks good and functional enough? Anything i could to fix or to optimize the code? Thanks in advance. I'm a huge noob so pardon me if i asked anything silly
解决方案
From my point the following implementation would be ok:
public static int getHammingDistance(String sequence1, String sequence2) {
    char[] s1 = sequence1.toCharArray();
    char[] s2 = sequence2.toCharArray();

    int shorter = Math.min(s1.length, s2.length);
    int longest = Math.max(s1.length, s2.length);

    int result = 0;
    for (int i=0; i<shorter; i++) {
        if (s1[i] != s2[i]) result++;
    }

    result += longest - shorter;

    return result;
}



uses array, what avoids the invocation of two method (charAt) for each single char that needs to be compared;
avoid exception when one string is longer than the other.


                        
这篇关于使用for循环来获得2个字符串之间的汉明距离的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！