除了一次次查还有别的变法么

public static int getCharacterPosition(String string,String substring,int count){
//这里是获取"/"符号的位置
Matcher slashMatcher = Pattern.compile(substring).matcher(string);
int mIdx = 0;
while(slashMatcher.find()) {
mIdx++;
//当"/"符号第三次出现的位置
if(mIdx == count){
break;
}
}
return slashMatcher.start();
}
已经找到并改写 ，大家可以参考

不用谢

public int xxx(int x,String str,String s){
int count=0;
for(int i=1;i<x;i++){
count=s.indexOf(str, int count);
}
return count;
}

public class Test {
/**
* @param args
*/
public static void main(String[] args) {
System.out.println(xxx(3,"a","saggasssags"));;
}
public static int xxx(int x,String str,String s){
int count=0;
for(int i=1;i<=x;i++){
count=s.indexOf(str,count+1);
}
return count;
}
}

public class Test {
/**
* @param args
*/
public static void main(String[] args) {
System.out.println(xxx(5, "a", "saggasssags"));
}
public static int xxx(int x, String str, String s) {
int count = 0;
for (int i = 1; i <= x; i++) {
count = s.indexOf(str, count + 1);
if (count == -1) {
return count;
}
}
return count;
}
}

这种问题，aaaa中查询aa到底算几个？2还是3？
如果是2那么ls的算法又出问题，我果然是坏人

public static int getIndex(String str, String s, int count) {
if (count == 0)
return -1;
int len = s.length();
int index = -1 - len;
while (count > 0 && (index = str.indexOf(s, index + len)) != -1) {
count--;
}
return index;
}
秀下限 @lx5266