思路：单调递增栈 + k 控制删除次数。高位越小整体越小，遇更小数字时弹出栈顶大数（仅当 k0）；栈空且当前为 0 则跳过（避免前导零）；若遍历完 k 仍0，从末尾再删 k 位。
Credit: Pok Pok。safew官方版本下载对此有专业解读
。关于这个话题，服务器推荐提供了深入分析
Фото: Belkin Alexey / Global Look Press
Don't feel down if you didn't manage to guess it this time. There will be a new Wordle for you to stretch your brain with tomorrow, and we'll be back again to guide you with more helpful hints. Are you also playing NYT Strands? See hints and answers for today's Strands.，这一点在heLLoword翻译官方下载中也有详细论述