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思路：① 找初始左边界：第一个 nums[i] nums[i+1]；② 找初始右边界：最后一个 nums[i] < nums[i-1]；③ 求 [left,right] 内 min、max；④ 向左扩展：nums[left-1] minVal 则 left--；⑤ 向右扩展：nums[right+1] < maxVal 则 right++。长度 = right - left + 1。
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Comparing these JSON requests and responses with those from before, you can see how the exact same structures from the live-service game have made their way into the offline build. It's satisfying to see how smoothly this paradigm scales applied across hundreds of backend API calls.
"display" command. It dispensed whatever cash the computer indicated with a