您的问题是，对于每个索引“i”，您都要检查从i开始到字符串结尾的子字符串。看一看https://www.w3schools.com/java/ref_string_indexof.asp
用这样的东西

String input = "B||B|B";
char bridgeMarker = 'B';
char spaceMarker = '|';

for (int i = 0; i < input.length(); i++) {
    if (input.charAt(i) == bridgeMarker) {
        System.out.println("B at index " + i);
    } else if (input.charAt(i) == spaceMarker) {
        System.out.println("| at index " + i);
    } else {
        System.out.println("Neither B nor | wos found");
    }
}

编辑要使用程序中的值，请将它们存储在列表中

String input = "B||B|B";
char bridgeMarker = 'B';
char spaceMarker = '|';

List<Integer> bridgeMarkers = new ArrayList<>();
List<Integer> spaceMarkers = new ArrayList<>();

for (int i = 0; i < input.length(); i++) {
    if (input.charAt(i) == bridgeMarker) {
        bridgeMarkers.add(i);
        System.out.println("B at index " + i);
    } else if (input.charAt(i) == spaceMarker) {
        spaceMarkers.add(i);
        System.out.println("| at index " + i);
    } else {
        System.out.println("Neither B nor | wos found");
    }
}

System.out.println("bridgeMarkers: "+Arrays.toString(bridgeMarkers.toArray()));
System.out.println("spaceMarkers: "+Arrays.toString(spaceMarkers.toArray()));

// how to use them
for (int i = 0; i < bridgeMarkers.size(); i++) {
    System.out.println("bridgeMarker nr " + i + " is at index " + bridgeMarkers.get(i));
}

for (int i = 0; i < spaceMarkers.size(); i++) {
    System.out.println("spaceMarker nr " + i + " is at index " + spaceMarkers.get(i));
}

通过使用java regex api，您可以更轻松地以不易出错的方式完成：

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Main {
    public static void main(String[] args) {
        String input = "B||B|B";
        Pattern pattern = Pattern.compile("[B\\|]");
        Matcher matcher = pattern.matcher(input);
        while (matcher.find()) {
            System.out.println(matcher.group() + " => " + matcher.start());
        }
    }
}

输出：

B => 0
| => 1
| => 2
B => 3
| => 4
B => 5

检查 Pattern 以及 Matcher 去了解他们。您还可以查看oracle的课程：正则表达式教程。