判断通过操作能否让字符串相等 I

You are given two strings s1 and s2, both of length 4, consisting of lowercase English letters.

You can apply the following operation on any of the two strings any number of times:

• Choose any two indices i and j such that j - i = 2, then swap the two characters at those indices in the string.

Return true if you can make the strings s1 and s2 equal, and false otherwise.

 

Example 1:

Input: s1 = "abcd", s2 = "cdab"
Output: true
Explanation: We can do the following operations on s1:
- Choose the indices i = 0, j = 2. The resulting string is s1 = "cbad".
- Choose the indices i = 1, j = 3. The resulting string is s1 = "cdab" = s2.

Example 2:

Input: s1 = "abcd", s2 = "dacb"
Output: false
Explanation: It is not possible to make the two strings equal.

 

Constraints:

• s1.length == s2.length == 4
• s1 and s2 consist only of lowercase English letters.

/*
 * 思路：
 * 利用Java Stream API，我们可以通过将字符串转换为字符流并检查相应的条件来实现相同的逻辑。
 * 通过对字符流进行过滤和匹配，可以验证是否存在符合交换条件的字符对。
 */
import java.util.stream.IntStream;

public class Solution {
    public boolean canTransform(String s1, String s2) {
        if (s1.equals(s2)) {
            return true;
        }
        return IntStream.range(0, 4)
                        .filter(i -> i % 2 == 0)
                        .allMatch(i -> s1.charAt(i) == s2.charAt(i + 2) && s1.charAt(i + 2) == s2.charAt(i)) &&
               IntStream.range(1, 4)
                        .filter(i -> i % 2 != 0)
                        .allMatch(i -> s1.charAt(i) == s2.charAt(i + 2) && s1.charAt(i + 2) == s2.charAt(i));
    }
}

这个题解的核心思路是分别比较字符串s1和s2中对应的可以通过特定操作（相隔两个位置的字符交换）交换的字符组。具体来说，s1和s2中的第一个字符与第三个字符可以互换，第二个字符与第四个字符可以互换。因此，题解检查了s1中的第一个字符和第三个字符是否能和s2中的第一个和第三个字符匹配（无论是原始顺序还是交换顺序），并且同样的检查对第二个和第四个字符进行了。只有两组字符都能匹配，函数才会返回true。

O(1)

O(1)