You are given a string s that contains some bracket pairs, with each pair containing a non-empty key.
"(name)is(age)yearsold", there are two bracket pairs that contain the keys "name" and "age".You know the values of a wide range of keys. This is represented by a 2D string array knowledge where each knowledge[i] = [keyi, valuei] indicates that key keyi has a value of valuei.
You are tasked to evaluate all of the bracket pairs. When you evaluate a bracket pair that contains some key keyi, you will:
keyi and the bracket pair with the key's corresponding valuei.keyi and the bracket pair with a question mark "?" (without the quotation marks).Each key will appear at most once in your knowledge. There will not be any nested brackets in s.
Return the resulting string after evaluating all of the bracket pairs.
\n\n\n
Example 1:
\n\n\nInput: s = "(name)is(age)yearsold", knowledge = [["name","bob"],["age","two"]]\nOutput: "bobistwoyearsold"\nExplanation:\nThe key "name" has a value of "bob", so replace "(name)" with "bob".\nThe key "age" has a value of "two", so replace "(age)" with "two".\n\n\n
Example 2:
\n\n\nInput: s = "hi(name)", knowledge = [["a","b"]]\nOutput: "hi?"\nExplanation: As you do not know the value of the key "name", replace "(name)" with "?".\n\n\n
Example 3:
\n\n\nInput: s = "(a)(a)(a)aaa", knowledge = [["a","yes"]]\nOutput: "yesyesyesaaa"\nExplanation: The same key can appear multiple times.\nThe key "a" has a value of "yes", so replace all occurrences of "(a)" with "yes".\nNotice that the "a"s not in a bracket pair are not evaluated.\n\n\n
\n
Constraints:
\n\n1 <= s.length <= 1050 <= knowledge.length <= 105knowledge[i].length == 21 <= keyi.length, valuei.length <= 10s consists of lowercase English letters and round brackets '(' and ')'.'(' in s will have a corresponding close bracket ')'.s will be non-empty.s.keyi and valuei consist of lowercase English letters.keyi in knowledge is unique.给你一个字符串 s ，它包含一些括号对，每个括号中包含一个 非空 的键。
\"(name)is(age)yearsold\" 中，有 两个 括号对，分别包含键 \"name\" 和 \"age\" 。你知道许多键对应的值，这些关系由二维字符串数组 knowledge 表示，其中 knowledge[i] = [keyi, valuei] ，表示键 keyi 对应的值为 valuei 。
你需要替换 所有 的括号对。当你替换一个括号对，且它包含的键为 keyi 时，你需要：
keyi 和括号用对应的值 valuei 替换。knowledge 中无法得知某个键对应的值，你需要将 keyi 和括号用问号 \"?\" 替换（不需要引号）。knowledge 中每个键最多只会出现一次。s 中不会有嵌套的括号。
请你返回替换 所有 括号对后的结果字符串。
\n\n\n\n
示例 1：
\n\n\n输入：s = \"(name)is(age)yearsold\", knowledge = [[\"name\",\"bob\"],[\"age\",\"two\"]]\n输出：\"bobistwoyearsold\"\n解释：\n键 \"name\" 对应的值为 \"bob\" ，所以将 \"(name)\" 替换为 \"bob\" 。\n键 \"age\" 对应的值为 \"two\" ，所以将 \"(age)\" 替换为 \"two\" 。\n\n\n
示例 2：
\n\n\n输入：s = \"hi(name)\", knowledge = [[\"a\",\"b\"]]\n输出：\"hi?\"\n解释：由于不知道键 \"name\" 对应的值，所以用 \"?\" 替换 \"(name)\" 。\n\n\n
示例 3：
\n\n\n输入：s = \"(a)(a)(a)aaa\", knowledge = [[\"a\",\"yes\"]]\n输出：\"yesyesyesaaa\"\n解释：相同的键在 s 中可能会出现多次。\n键 \"a\" 对应的值为 \"yes\" ，所以将所有的 \"(a)\" 替换为 \"yes\" 。\n注意，不在括号里的 \"a\" 不需要被替换。\n\n\n
\n\n
提示：
\n\n1 <= s.length <= 1050 <= knowledge.length <= 105knowledge[i].length == 21 <= keyi.length, valuei.length <= 10s 只包含小写英文字母和圆括号 '(' 和 ')' 。s 中每一个左圆括号 '(' 都有对应的右圆括号 ')' 。s 中每对括号内的键都不会为空。s 中不会有嵌套括号对。keyi 和 valuei 只包含小写英文字母。knowledge 中的 keyi 不会重复。\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 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\\u4e00\\u4e9b\\u5e38\\u7528\\u7684\\u6570\\u636e\\u7ed3\\u6784\\u5df2\\u5728 Algorithms \\u6a21\\u5757\\u4e2d\\u63d0\\u4f9b\\uff1ahttps:\\/\\/www.rubydoc.info\\/github\\/kanwei\\/algorithms\\/Algorithms<\\/p>\"],\"swift\":[\"Swift\",\" \\u7248\\u672c\\uff1a \\u6211\\u4eec\\u901a\\u5e38\\u4fdd\\u8bc1\\u66f4\\u65b0\\u5230 Apple\\u653e\\u51fa\\u7684\\u6700\\u65b0\\u7248Swift<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u53d1\\u73b0Swift\\u4e0d\\u662f\\u6700\\u65b0\\u7248\\u7684\\uff0c\\u8bf7\\u8054\\u7cfb\\u6211\\u4eec\\uff01\\u6211\\u4eec\\u5c06\\u5c3d\\u5feb\\u66f4\\u65b0\\u3002<\\/p>\"],\"golang\":[\"Go\",\" \\u7248\\u672c\\uff1a \\u652f\\u6301 https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods@v1.18.1<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982array<\\/a>, bisect<\\/a>, 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