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(); } }帮我写完整的代码

这是一个去除字符串中重复字母的算法，下面是完整的代码实现：

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();
    }
}

该算法的具体思路是用一个栈来维护一个字典序最小的字符串，并且保证每个字符只出现一次。对于每个字符，先判断它是否已经在栈中出现过，如果出现过，那么不需要考虑，直接跳过；否则，需要判断栈顶元素和当前元素的大小关系，如果栈顶元素大于当前元素且栈顶元素在后面还会出现，那么就需要将栈顶元素弹出，然后再判断新的栈顶元素和当前元素的大小关系，直到栈顶元素小于等于当前元素为止。最后将栈中元素弹出并倒序输出即可。