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。
"It's a huge surprise because I've always been able to get a job very easily," said Trigg, who lives in Texas. "It wasn't even on my radar to be prepared for more than six months of unemployment. It wasn't in my universe."
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