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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。
。业内人士推荐搜狗输入法2026作为进阶阅读
Are you also playing NYT Strands? See hints and answers for today's Strands.，推荐阅读heLLoword翻译官方下载获取更多信息
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getChunks() { return chunks; }