You are given a 0-indexed string pattern of length n consisting of the characters 'I' meaning increasing and 'D' meaning decreasing.
A 0-indexed string num of length n + 1 is created using the following conditions:
num consists of the digits '1' to '9', where each digit is used at most once.pattern[i] == 'I', then num[i] < num[i + 1].pattern[i] == 'D', then num[i] > num[i + 1].Return the lexicographically smallest possible string num that meets the conditions.
\n
Example 1:
\n\n\nInput: pattern = "IIIDIDDD"\nOutput: "123549876"\nExplanation:\nAt indices 0, 1, 2, and 4 we must have that num[i] < num[i+1].\nAt indices 3, 5, 6, and 7 we must have that num[i] > num[i+1].\nSome possible values of num are "245639871", "135749862", and "123849765".\nIt can be proven that "123549876" is the smallest possible num that meets the conditions.\nNote that "123414321" is not possible because the digit '1' is used more than once.\n\n
Example 2:
\n\n\nInput: pattern = "DDD"\nOutput: "4321"\nExplanation:\nSome possible values of num are "9876", "7321", and "8742".\nIt can be proven that "4321" is the smallest possible num that meets the conditions.\n\n\n
\n
Constraints:
\n\n1 <= pattern.length <= 8pattern consists of only the letters 'I' and 'D'.给你下标从 0 开始、长度为 n 的字符串 pattern ，它包含两种字符，'I' 表示 上升 ，'D' 表示 下降 。
你需要构造一个下标从 0 开始长度为 n + 1 的字符串，且它要满足以下条件：
num 包含数字 '1' 到 '9' ，其中每个数字 至多 使用一次。pattern[i] == 'I' ，那么 num[i] < num[i + 1] 。pattern[i] == 'D' ，那么 num[i] > num[i + 1] 。请你返回满足上述条件字典序 最小 的字符串 num。
\n\n
示例 1：
\n\n\n输入：pattern = \"IIIDIDDD\"\n输出：\"123549876\"\n解释：\n下标 0 ，1 ，2 和 4 处，我们需要使 num[i] < num[i+1] 。\n下标 3 ，5 ，6 和 7 处，我们需要使 num[i] > num[i+1] 。\n一些可能的 num 的值为 \"245639871\" ，\"135749862\" 和 \"123849765\" 。\n\"123549876\" 是满足条件最小的数字。\n注意，\"123414321\" 不是可行解因为数字 '1' 使用次数超过 1 次。\n\n
示例 2：
\n\n\n输入：pattern = \"DDD\"\n输出：\"4321\"\n解释：\n一些可能的 num 的值为 \"9876\" ，\"7321\" 和 \"8742\" 。\n\"4321\" 是满足条件最小的数字。\n\n\n
\n\n
提示：
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"exampleTestcases": "\"IIIDIDDD\"\n\"DDD\"",
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