LeetCode 1487. 保证文件名唯一（哈希map）

示例 1：
输入：names = ["pes","fifa","gta","pes(2019)"]
输出：["pes","fifa","gta","pes(2019)"]
解释：文件系统将会这样创建文件名：
"pes" --> 之前未分配，仍为 "pes"
"fifa" --> 之前未分配，仍为 "fifa"
"gta" --> 之前未分配，仍为 "gta"
"pes(2019)" --> 之前未分配，仍为 "pes(2019)"

示例 2：
输入：names = ["gta","gta(1)","gta","avalon"]
输出：["gta","gta(1)","gta(2)","avalon"]
解释：文件系统将会这样创建文件名：
"gta" --> 之前未分配，仍为 "gta"
"gta(1)" --> 之前未分配，仍为 "gta(1)"
"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)" 。
"avalon" --> 之前未分配，仍为 "avalon"

示例 3：
输入：names = ["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece"]
输出：["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece(4)"]
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)"。

示例 4：
输入：names = ["wano","wano","wano","wano"]
输出：["wano","wano(1)","wano(2)","wano(3)"]
解释：每次创建文件夹 "wano" 时，只需增加后缀中 k 的值即可。

示例 5：
输入：names = ["kaido","kaido(1)","kaido","kaido(1)"]
输出：["kaido","kaido(1)","kaido(2)","kaido(1)(1)"]
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) 。
 
提示：
1 <= names.length <= 5 * 10^4
1 <= names[i].length <= 20
names[i] 由小写英文字母、数字和/或圆括号组成。

class Solution {	//C++
public:
    vector<string> getFolderNames(vector<string>& names) {
    	unordered_map<string,int> m;
    	vector<string> ans(names.size());
        int num;
    	for(int i = 0; i < names.size(); ++i)
    	{
    		if(!m.count(names[i]))//没有出现过
    		{
    			m[names[i]]++;//计数+1
    			ans[i] = names[i];//答案不变
    		}
    		else
    		{	//出现过该单词
                num = m[names[i]];//读取当前应该的计数
    			while(m.count(names[i]+"("+to_string(num)+")"))//出现过了
                    num++;//数字增加，直到没有出现过
                ans[i] = names[i]+"("+to_string(num)+")";//写入答案
    			m[names[i]] = num+1;//更新计数
    			m[ans[i]]++;
    		}
    	}
    	return ans;
    }
};

class Solution:# py3
    def getFolderNames(self, names: List[str]) -> List[str]:
        m = {}
        ans = [""]*len(names)
        for i in range(len(names)):
            if names[i] not in m:
                m[names[i]] = 1
                ans[i] = names[i]
            else:
                num = m[names[i]]
                while names[i]+'('+str(num)+')' in m:
                    num += 1
                ans[i] = names[i]+'('+str(num)+')'
                m[names[i]] = num+1
                m[ans[i]] = 1
        return ans