给我写出完整的可运行的代码public class Solution { public String removeDuplicateLetters(String s) { int[] tem = new int[26]; for (int i = 0; i < s.length(); i++) { char tems = s.charAt(i); tem[tems - 'a']++; } boolean[] visited = new boolean[26]; Stack<Character> stack = new Stack<>(); for (int i = 0; i < s.length(); i++) { char get = s.charAt(i); while (!stack.isEmpty() && stack.peek() > get && !visited[get - 'a'] && tem[stack.peek() - 'a'] > 0) { visited[stack.peek() - 'a'] = false; stack.pop(); } if (visited[get - 'a']) { tem[get - 'a']--; continue; } visited[get - 'a'] = true; stack.push(get); tem[get - 'a']--; } StringBuilder sBuilder = new StringBuilder(); while (!stack.isEmpty()) { sBuilder.append(stack.pop()); } return sBuilder.reverse().toString(); }的 }

这里是完整可运行的代码，您可以直接复制粘贴并运行：

import java.util.*;

public class Solution {
    public String removeDuplicateLetters(String s) {
        int[] tem = new int[26];
        for (int i = 0; i < s.length(); i++) {
            char tems = s.charAt(i);
            tem[tems - 'a']++;
        }
        boolean[] visited = new boolean[26];
        Stack<Character> stack = new Stack<>();
        for (int i = 0; i < s.length(); i++) {
            char get = s.charAt(i);
            while (!stack.isEmpty() && stack.peek() > get && !visited[get - 'a'] && tem[stack.peek() - 'a'] > 0) {
                visited[stack.peek() - 'a'] = false;
                stack.pop();
            }
            if (visited[get - 'a']) {
                tem[get - 'a']--;
                continue;
            }
            visited[get - 'a'] = true;
            stack.push(get);
            tem[get - 'a']--;
        }
        StringBuilder sBuilder = new StringBuilder();
        while (!stack.isEmpty()) {
            sBuilder.append(stack.pop());
        }
        return sBuilder.reverse().toString();
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        String s = "bcabc";
        String result = solution.removeDuplicateLetters(s);
        System.out.println(result);
    }
}

在这个示例中，我们调用了removeDuplicateLetters方法来移除字符串"bcabc"中的重复字符，并将结果打印到控制台上。