859. Buddy Strings-亲密字符串

Given two strings A and B of lowercase letters, return true if you can swap two letters in A so the result is equal to B, otherwise, return false.

Swapping letters is defined as taking two indices i and j (0-indexed) such that i != j and swapping the characters at A[i] and A[j]. For example, swapping at indices 0 and 2 in "abcd" results in "cbad".

Example 1:
Input: A = "ab", B = "ba"
Output: true
Explanation: You can swap A[0] = ‘a’ and A[1] = ‘b’ to get "ba", which is equal to B.

Example 2:
Input: A = "ab", B = "ab"
Output: false
Explanation: The only letters you can swap are A[0] = ‘a’ and A[1] = ‘b’, which results in "ba" != B.

Example 3:
Input: A = "aa", B = "aa"
Output: true
Explanation: You can swap A[0] = ‘a’ and A[1] = ‘a’ to get "aa", which is equal to B.

Example 4:
Input: A = "aaaaaaabc", B = "aaaaaaacb"
Output: true

Example 5:
Input: A = "", B = "aa"
Output: false

Constraints:

• 0 <= A.length <= 20000
• 0 <= B.length <= 20000
• A and B consist of lowercase letters.

Solution in python:

class Solution:
    def buddyStrings(self, A: str, B: str) -> bool:
        if len(A) != len(B):
            return False
        count = 0
        alist = []
        blist = []
        for i in range(len(A)):
            if A[i] != B[i]:
                count += 1
                alist.append(A[i])
                blist.append(B[i])
            if count > 2:
                return False
        if count == 2 and alist[0] == blist[1] and alist[1] == blist[0]:
            return True
        if count == 0 and len(A) != len(set(A)):
            return True
        return False