技術標籤：LeetCode

題目：

Two strings are consideredcloseif you can attain one from the other using the following operations:

• Operation 1: Swap any twoexistingcharacters.
  • For example,abcde-> aecdb
• Operation 2: Transformeveryoccurrence of oneexistingcharacter into anotherexistingcharacter, and do the same with the other character.
  • For example,aa
    cabb->bbcbaa(alla's turn intob's, and allb's turn intoa's)

You can use the operations on either string as many times as necessary.

Given two strings,word1andword2, returntrueifword1andword2areclose, andfalseotherwise.

Example 1:

Input: word1 = "abc", word2 = "bca"
Output: true
Explanation:
 
 You can attain word2 from word1 in 2 operations.
Apply Operation 1: "abc" -> "acb"
Apply Operation 1: "acb" -> "bca"

Example 2:

Input: word1 = "a", word2 = "aa"
Output: false
Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.

Example 3:

Input: word1 = "cabbba", word2 = "abbccc"
Output: true
Explanation: You can attain word2 from word1 in 3 operations.
Apply Operation 1: "cabbba" -> "caabbb"
Apply Operation 2: "caabbb" -> "baaccc"
Apply Operation 2: "baaccc" -> "abbccc"

Example 4:

Input: word1 = "cabbba", word2 = "aabbss"
Output: false
Explanation: It is impossible to attain word2 from word1, or vice versa, in any amount of operations.

Constraints:

• 1 <= word1.length, word2.length <= 10^5
• word1andword2containonly lowercase English letters.

思路：

這道題看上去有點花裡胡哨，但是大概是Contest第二題的樣子，所有理論上不會太難。讀完題目我們可以用三個條件來制約給定的兩個陣列，以此判斷他們是否close。首先長度肯定要一樣；其次因為我們關心的是字元以及字元出現的次數，順序無關緊要，所以我們可以sort。對於出現的字元，我們set來記錄；而字元出現的次數，因為是一個一直要改變的key（每次出現已出現字元，次數需要加一），因此用set就不太合適，但是如果用hash map，在比較的時候又不好sort（我們只需要sort次數，而對應次數是什麼字元我們並不在意），同時考慮到總共也就只有26個字母，可以直接用vector來記錄頻次，只要用當前字元減去'a'就是當前字元在vector中的index--有一點點像簡易版hashmap。最後只要保證sort完的兩個vector和set都相同即可。

程式碼：

class Solution {
public:
bool closeStrings(string word1, string word2) {
int len1=word1.size();
int len2=word2.size();
if(len1!=len2)
return 0;
set<char> s1, s2;
vector<int> fre1(26), fre2(26);
for(auto c:word1)
{
s1.insert(c);
fre1[c-'a']++;
}
for(auto c:word2)
{
s2.insert(c);
fre2[c-'a']++;
}
sort(fre1.begin(),fre1.end());
sort(fre2.begin(),fre2.end());
return s1==s2&&fre1==fre2;
}
};