Given an array of strings names of size n. You will create n folders in your file system such that, at the ith minute, you will create a folder with the name names[i].
Since two files cannot have the same name, if you enter a folder name that was previously used, the system will have a suffix addition to its name in the form of (k), where, k is the smallest positive integer such that the obtained name remains unique.
Return an array of strings of length n where ans[i] is the actual name the system will assign to the ith folder when you create it.
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Example 1:
\n\n\nInput: names = ["pes","fifa","gta","pes(2019)"]\nOutput: ["pes","fifa","gta","pes(2019)"]\nExplanation: Let's see how the file system creates folder names:\n"pes" --> not assigned before, remains "pes"\n"fifa" --> not assigned before, remains "fifa"\n"gta" --> not assigned before, remains "gta"\n"pes(2019)" --> not assigned before, remains "pes(2019)"\n\n\n
Example 2:
\n\n\nInput: names = ["gta","gta(1)","gta","avalon"]\nOutput: ["gta","gta(1)","gta(2)","avalon"]\nExplanation: Let's see how the file system creates folder names:\n"gta" --> not assigned before, remains "gta"\n"gta(1)" --> not assigned before, remains "gta(1)"\n"gta" --> the name is reserved, system adds (k), since "gta(1)" is also reserved, systems put k = 2. it becomes "gta(2)"\n"avalon" --> not assigned before, remains "avalon"\n\n\n
Example 3:
\n\n\nInput: names = ["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece"]\nOutput: ["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece(4)"]\nExplanation: When the last folder is created, the smallest positive valid k is 4, and it becomes "onepiece(4)".\n\n\n
\n
Constraints:
\n\n1 <= names.length <= 5 * 1041 <= names[i].length <= 20names[i] consists of lowercase English letters, digits, and/or round brackets.给你一个长度为 n 的字符串数组 names 。你将会在文件系统中创建 n 个文件夹：在第 i 分钟，新建名为 names[i] 的文件夹。
由于两个文件 不能 共享相同的文件名，因此如果新建文件夹使用的文件名已经被占用，系统会以 (k) 的形式为新文件夹的文件名添加后缀，其中 k 是能保证文件名唯一的 最小正整数 。
返回长度为 n 的字符串数组，其中 ans[i] 是创建第 i 个文件夹时系统分配给该文件夹的实际名称。
\n\n
示例 1：
\n\n输入：names = ["pes","fifa","gta","pes(2019)"]\n输出：["pes","fifa","gta","pes(2019)"]\n解释：文件系统将会这样创建文件名：\n"pes" --> 之前未分配，仍为 "pes"\n"fifa" --> 之前未分配，仍为 "fifa"\n"gta" --> 之前未分配，仍为 "gta"\n"pes(2019)" --> 之前未分配，仍为 "pes(2019)"\n\n\n
示例 2：
\n\n输入：names = ["gta","gta(1)","gta","avalon"]\n输出：["gta","gta(1)","gta(2)","avalon"]\n解释：文件系统将会这样创建文件名：\n"gta" --> 之前未分配，仍为 "gta"\n"gta(1)" --> 之前未分配，仍为 "gta(1)"\n"gta" --> 文件名被占用，系统为该名称添加后缀 (k)，由于 "gta(1)" 也被占用，所以 k = 2 。实际创建的文件名为 "gta(2)" 。\n"avalon" --> 之前未分配，仍为 "avalon"\n\n\n
示例 3：
\n\n输入：names = ["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece"]\n输出：["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece(4)"]\n解释：当创建最后一个文件夹时，最小的正有效 k 为 4 ，文件名变为 "onepiece(4)"。\n\n\n
示例 4：
\n\n输入：names = ["wano","wano","wano","wano"]\n输出：["wano","wano(1)","wano(2)","wano(3)"]\n解释：每次创建文件夹 "wano" 时，只需增加后缀中 k 的值即可。\n\n
示例 5：
\n\n输入：names = ["kaido","kaido(1)","kaido","kaido(1)"]\n输出：["kaido","kaido(1)","kaido(2)","kaido(1)(1)"]\n解释：注意，如果含后缀文件名被占用，那么系统也会按规则在名称后添加新的后缀 (k) 。\n\n\n
\n\n
提示：
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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 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. 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