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 which is 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.
Example 1:
Input: names = ["pes","fifa","gta","pes(2019)"] Output: ["pes","fifa","gta","pes(2019)"] Explanation: Let's see how the file system creates folder names: "pes" --> not assigned before, remains "pes" "fifa" --> not assigned before, remains "fifa" "gta" --> not assigned before, remains "gta" "pes(2019)" --> not assigned before, remains "pes(2019)"Example 2:
Input: names = ["gta","gta(1)","gta","avalon"] Output: ["gta","gta(1)","gta(2)","avalon"] Explanation: Let's see how the file system creates folder names: "gta" --> not assigned before, remains "gta" "gta(1)" --> not assigned before, remains "gta(1)" "gta" --> the name is reserved, system adds (k), since "gta(1)" is also reserved, systems put k = 2. it becomes "gta(2)" "avalon" --> not assigned before, remains "avalon"Example 3:
Input: names = ["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece"] Output: ["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece(4)"] Explanation: When the last folder is created, the smallest positive valid k is 4, and it becomes "onepiece(4)".Example 4:
Input: names = ["wano","wano","wano","wano"] Output: ["wano","wano(1)","wano(2)","wano(3)"] Explanation: Just increase the value of k each time you create folder "wano".Example 5:
Input: names = ["kaido","kaido(1)","kaido","kaido(1)"] Output: ["kaido","kaido(1)","kaido(2)","kaido(1)(1)"] Explanation: Please note that system adds the suffix (k) to current name even it contained the same suffix before.
Constraints:
1 <= names.length <= 5 * 10^41 <= names[i].length <= 20names[i] consists of lower case English letters, digits and/or round brackets.题解：
使用一个hashmap存储每个字符串出现的序号。
class Solution { public String[] getFolderNames(String[] names) { String[] res = new String[names.length]; HashMap<String, Integer> hashMap = new HashMap<String, Integer>(); for (int i = 0; i < names.length; i++) { if (hashMap.containsKey(names[i]) == true) { int num = hashMap.get(names[i]); while (hashMap.containsKey(names[i] + "(" + num + ")") == true) { num++; } hashMap.put(names[i] + "(" + num + ")", 1); hashMap.put(names[i], hashMap.get(names[i]) + 1); res[i] = names[i] + "(" + num + ")"; } else { hashMap.put(names[i], 1); res[i] = names[i]; } } return res; } }
