我在初学java，我想通过字符在Java焦炭两个字符串比较，发现他们有多少不同的字符具有由下面的代码，但它不工作，
min is the min between the 2 strings
for(int i=0; i<min-1; i++){
s1 = w1.substring(j,j++);
s2 = w2.substring(j,j++);
if (! s1.equalsIgnoreCase(s2) ){
counter++;
}
}`
有小费吗？
用这个：
char[] first = w1.toLowerCase().toCharArray();
char[] second = w2.toLowerCase().toCharArray();
int minLength = Math.min(first.length, second.length);
for(int i = 0; i < minLength; i++)
{
if (first[i] != second[i])
{
counter++;
}
}
使用的charAt（index）方法，并使用“==”操作两个字符：
c1 = w1.charAt(j);
c2 = w2.charAt(j);
if (c1 == c2) ){
counter++;
}
int i =0;
for(char c : w1.toCharArray())){
if(i < w2.length() && w2.charAt(i++) != c)
counter++;
}
我们可以解决这个问题substring 。 但是，让我们看看你的代码首先：
// assuming, min is the minimum length of both strings,
// then you don't check the char at the last position
for(int j=0; j < min-1; j++) {
// s1, s2 will always be empty strings, because j++ is post-increment:
// it will be incremented *after* it has been evaluated
s1 = w1.substring(j,j++);
s2 = w2.substring(j,j++);
if (!s1.equalsIgnoreCase(s2) ){
counter++;
}
}
基于溶液substring可能是这样的：
for(int j=0; j < min; j++) {
s1 = w1.substring(j,j+1);
s2 = w2.substring(j,j+1);
if (!s1.equalsIgnoreCase(s2) ){
counter++;
}
}
从java培训教程我的笔记需要用的charAt（）和嵌套循环字符串比较...的方法可以很容易地修改，从源字符串返回不匹配的字符...但我会离开，一个由你... ;-)
public class SubString {
public static boolean findTarget( String target, String source ) {
int target_len = target.length();
int source_len = source.length();
boolean found = false;
for(int i = 0; ( i < source_len && !found ); ++i) {
int j = 0;
while( !found ) {
if( j >= target_len ) {
break;
}
/**
* Learning Concept:
*
* String target = "for";
* String source = "Searching for a string within a string the hard way.";
*
* 1 - target.charAt( j ) :
* The character at position 0 > The first character in 'Target' > character 'f', index 0.
*
* 2 - source.charAt( i + j) :
*
* The source strings' array index is searched to determine if a match is found for the
* target strings' character index position. The position for each character in the target string
* is then compared to the position of the character in the source string.
*
* If the condition is true, the target loop continues for the length of the target string.
*
* If all of the source strings' character array element position matches the target strings' character array element position
* Then the condition succeeds ..
*/
else if( target.charAt( j ) != source.charAt( i + j ) ) {
break;
} else {
++j;
if( j == target_len ) {
found = true;
}
}
}
}
return found;
}
public static void main ( String ... args ) {
String target = "for";
String source = "Searching for a string within a string the hard way.";
System.out.println( findTarget(target, source) );
}
}
