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 * 10^4 -
1 <= names[i].length <= 20 -
names[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;
}
}
