Что думаешь? Оцени!
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。
，这一点在夫子中也有详细论述
They expanded their leaked-credential detection pipeline to cover the keys we reported, thereby proactively protecting real Google customers from threat actors exploiting their Gemini API keys. They also committed to fixing the root cause, though we haven't seen a concrete outcome yet.，更多细节参见91视频
以“中国样本”赋能“区域健康”
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