文章目录
一、题目
1657. Determine if Two Strings Are Close
Two strings are considered close if you can attain one from the other using the following operations:
Operation 1: Swap any two existing characters.
For example, abcde -> aecdb
Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
For example, aacabb -> bbcbaa (all a's turn into b's, and all b's turn into a's)
You can use the operations on either string as many times as necessary.
Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.
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"
Constraints:
1 <= word1.length, word2.length <= 105
word1 and word2 contain only lowercase English letters.
题目来源： leetcode
二、题解
当两个字符串，所拥有的共同字符类型完全相同，且字母出现数目以及出现该数目的个数完全相同时，这两个字符串是close的。
cpp
class Solution {
public:
bool closeStrings(string word1, string word2) {
int n1 = word1.length();
int n2 = word2.length();
vector<int> map1(26,0);
vector<int> map2(26,0);
vector<int> times(max(n1,n2) + 1,0);
for(int i = 0;i < n1;i++) map1[word1[i] - 'a']++;
for(int i = 0;i < n2;i++) map2[word2[i] - 'a']++;
//如果有字母不在交集中
for(int i = 0;i < 26;i++){
if((map1[i] == 0 && map2[i] != 0) || (map1[i] != 0 && map2[i] == 0)) return false;
}
//统计出现次数的个数
for(int i = 0;i < 26;i++){
if(map1[i] != 0) {
times[map1[i]]++;
}
}
for(int i = 0;i < 26;i++){
if(map2[i] != 0) times[map2[i]]--;
if(times[map2[i]] < 0) return false;
}
return true;
}
};