You are given a string s and a robot that currently holds an empty string t. Apply one of the following operations until s and t are both empty:

Remove the first character of a string s and give it to the robot. The robot will append this character to the string t.
Remove the last character of a string t and give it to the robot. The robot will write this character on paper.
Return the lexicographically smallest string that can be written on the paper.

簡單來說我們有一個字串s和空字串t，然後有兩個行爲可以做：

1. 把s字串的第一個字符移動到t
2. 把t字串的最後一個字符移出到紙上
   重複前兩個行為直到s,t都變成空字串，目標是返回紙上的字（盡可能字典序最小的字串）
   老實說題目翻譯過後我依然沒有很懂，所以直接上題目範例比較好理解：

• Example :
  Input: s = "bdda"
  Output: "addb"
  Explanation: Let p denote the written string.
  Initially p="", s="bdda", t="".
  Perform first operation four times p="", s="", t="bdda".
  Perform second operation four times p="addb", s="", t="".
  字典序的意思就是abcd...這樣的排列

我過了很久都想不到該怎麼解，
加上我的偏頭痛又發作了，最終決定求助chatgpt...我只負責了筆記註釋
它似乎是使用了模擬的方法
（用Stack來模擬字串t，堆疊符合後進先出的原則，適合模擬字符的添加和移除。）

• stack.peek()＝字符字典序

class Solution {
    public String robotWithString(String s) {
        // 存放t的堆疊
        Stack<Character> stack = new Stack<>();
        // 字串結果
        StringBuilder result = new StringBuilder();
        
        // 將s轉換成字元陣列，方便操作！
        char[] chars = s.toCharArray();
        // 儲存每個字元右邊最小的字元
        char[] rightMin = new char[chars.length];
        
        // 計算每個位置後方的最小字母，方便後續比對
        rightMin[chars.length - 1] = chars[chars.length - 1];
        for (int i = chars.length - 2; i >= 0; i--) {
            rightMin[i] = (char) Math.min(chars[i], rightMin[i + 1]);
        }
        
        // 遍歷字串s，模擬將字母加入t
        int i = 0;
        while (i < chars.length || !stack.isEmpty()) {
            // 當可以從s取字元並放入堆疊t時
            if (i < chars.length && (stack.isEmpty() || stack.peek() > rightMin[i])) {
                stack.push(chars[i]);
                i++;
            } else {
                // 如果棧中字符比目前s中的字母字典序小，則取出並放到結果中
                result.append(stack.pop());
            }
        }
        return result.toString();
    }
}

成功。