java 使用for循环获取2个字符串之间的汉明距离
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Using for loop to get the Hamming distance between 2 strings
提问by Doh
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.
在这个任务中，我需要得到两个字符串序列 1 和序列 2 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对应符号不同的位置数 - 来自维基百科）。
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.
首先，我制作了 2 根新弦，这是 2 根原始弦，但都采用了小写字母，以便于比较。然后我求助于使用 for 循环和 if 来比较 2 个字符串。对于这 2 对字符串中字符的任何差异，循环会将 1 添加到 int x = 0。该方法的返回将是此 x 的值。
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
那么代码看起来不错且功能足够吗？我可以修复或优化代码吗？提前致谢。我是个大菜鸟，如果我问什么傻话，请原谅我
回答by Francisco Spaeth
From my point the following implementation would be ok:
从我的角度来看，以下实现是可以的：
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.
- 使用数组，避免为每个需要比较的单个字符调用两个方法（charAt）；
- 当一个字符串比另一个长时避免异常。
回答by radai
your code is completely off. as you said yourself, the distance is the number of places where the strings differ - so you should only have 1 loop, going over both strings at once. instead you have 2 nested loops that compare every index in string a to every index in string b.
您的代码完全关闭。正如您自己所说，距离是字符串不同的地方的数量 - 所以你应该只有 1 个循环，一次遍历两个字符串。相反，您有 2 个嵌套循环，将字符串 a 中的每个索引与字符串 b 中的每个索引进行比较。
also, writing an if condition that results in a+=0is a waste of time.
此外，编写导致结果的 if 条件a+=0是浪费时间。
try this instead:
试试这个：
for (int x = 0; x < sequenceX.length(); x++) { //both are of the same length
if (sequenceX.charAt(x) != sequenceY.charAt(x)) {
a += 1;
}
}
also, this is still a naive approach which will probbaly not work with complex unicode characters (where 2 characters can be logically equal yet not have the same character code)
此外，这仍然是一种天真的方法，可能不适用于复杂的 unicode 字符（其中 2 个字符在逻辑上可以相等但不具有相同的字符代码）
回答by threadfin
public static int getHammingDistance(String sequenceX, String sequenceY) {
int a = 0;
// String sequenceX = sequence1.toLowerCase();
//String sequenceY = sequence2.toLowerCase();
if (sequenceX.length() != sequenceY.length()) {
return -1; //input strings should be of equal length
}
for (int i = 0; i < sequenceX.length(); i++) {
if (sequenceX.charAt(i) != sequenceY.charAt(i)) {
a++;
}
}
return a;
}
回答by AlexR
Your code is OK, however I'd suggest you the following improvements.
您的代码没问题，但是我建议您进行以下改进。
- do not use
charAt()of string. Get char array from string usingtoCharArray()before loop and then work with this array. This is more readable and more effective. The structure
if (sequenceX.charAt(x) == sequenceY.charAt(y)) { a += 0; } else if (sequenceX.charAt(x) != sequenceY.charAt(y)) { a += 1; }looks redundant. Fix it to: if (sequenceX.charAt(x) == sequenceY.charAt(y)) { a += 0; } else { a += 1; }
- 不要使用
charAt()字符串。使用toCharArray()before 循环从字符串中获取字符数组，然后使用此数组。这更具可读性和更有效。 结构
if (sequenceX.charAt(x) == sequenceY.charAt(y)) { a += 0; } else if (sequenceX.charAt(x) != sequenceY.charAt(y)) { a += 1; }看起来多余。将其修复为： if (sequenceX.charAt(x) == sequenceY.charAt(y)) { a += 0; } else { a += 1; }
Moreover taking into account that I recommended you to work with array change it to something like:
此外，考虑到我建议您使用数组将其更改为以下内容：
a += seqx[x] == seqY[x] ? 0 : 1
a += seqx[x] == seqY[x] ? 0 : 1
less code less bugs...
更少的代码更少的错误...
EDIT: as mentionded by @radai you do not need if/elsestructure at all: adding 0to ais redundant.
编辑：正如@radai 所提到的，您根本不需要if/else结构：添加0到a是多余的。
