Determine if Two Strings Are Close (M)

题目

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.

题意

按指定规则判断两个字符串是否近似。

思路

根据规则可以得到三点信息：

1. 两字符串长度必须相等；
2. 两字符串中出现的字母种类必须相同；
3. 两字符串中所有字母的出现次数排序后必须相同。

代码实现

Java

class Solution {
    public boolean closeStrings(String word1, String word2) {
        if (word1.length() != word2.length()) {
            return false;
        }

        int[] hash1 = new int[26], hash2 = new int[26];
        for (int i = 0; i < word1.length(); i++) {
            hash1[word1.charAt(i) - 'a']++;
            hash2[word2.charAt(i) - 'a']++;
        }

        for (int i = 0; i < 26; i++) {
            if (hash1[i] > 0 && hash2[i] == 0 || hash1[i] == 0 && hash2[i] > 0) {
                return false;
            }
        }

        Arrays.sort(hash1);
        Arrays.sort(hash2);

        for (int i = 0; i < 26; i++) {
            if (hash1[i] != hash2[i]) {
                return false;
            }
        }

        return true;
    }
}

标签：operations,hash2,1657,hash1,word1,word2,Close,Operation,Strings
来源： https://www.cnblogs.com/mapoos/p/14316850.html