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LeetCode 1657. Determine if Two Strings Are Close

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题目

原题在此

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”

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 <= $10^5$
  • word1 and word2 contain only lowercase English letters.

解析

根据题意, close的条件如下:

  1. 出现过的字符组成的集合(不重复)相同. 以例三为例: set<char> word1 = {a, b, c}, word2 = {a, b, c}
  2. 统计每个字符出现过的次数所组成的集合(可重复)相同. 以例三为例: multiset<int> word1 = {2, 3, 1}, word2 = {1, 2, 3}
  3. 因为题目指出输入字符串中只含有小写字母, 可使用数组替换集合.

代码

std::set使用

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class Solution {
public:
    bool closeStrings(string word1, string word2) {
        set<int> char1, char2;
        multiset<int> count1, count2;
        map<int, int> m1, m2;
        for(char c : word1) m1[c - 'a'] ++;
        for(char c : word2) m2[c - 'a'] ++;
        
        map<int, int>::iterator it;
        for (it = m1.begin(); it != m1.end(); it++) {
            char1.insert(it -> first);
            count1.insert(it -> second);
        }

        for (it = m2.begin(); it != m2.end(); it++) {
            char2.insert(it -> first);
            count2.insert(it -> second);
        }
        
        return char1 == char2 && count1 == count2;
    }
};

std::set不使用

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class Solution {
public:
    bool closeStrings(string word1, string word2) {
        array<int, 26> m1 {};
        array<int, 26> m2 {};
        for(char c : word1) m1[c - 'a'] ++;
        for(char c : word2) m2[c - 'a'] ++;
        
        if (!equal(begin(m1), end(m1), begin(m2), end(m2), [](int a, int b) { return bool (a) == bool (b); }))
            return false;
        // 此处sort数组长度为26和输入数据长度无关
        sort(begin(m1), end(m1));
        sort(begin(m2), end(m2));
        return m1 == m2;
    }
};