題目：

Given two strings A and B of lowercase letters, return true if and only if we can swap two letters in A so that the result equals B.
Example 1:

Input: A = "ab", B = "ba" 
Output: true 

Example 2:

Input: A = "ab", B = "ab" 
Output: false 

Example 3:

Input: A = "aa", B = "aa" 
Output: true 
 

Example 4:

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

Example 5:

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

Note:
0 <= A.length <= 20000 0 <= B.length <= 20000
A and B consist only of lowercase letters.

解釋：
如果A和B長度不相等，一定是false
如果A和B相等，那麼A中一定要有重複元素，而且交換的時候交換重複元素才能true。
如果A和B僅僅是長度相等而本身不相等，那麼順序遍歷，若A[i]和B[i]不相等，則記錄這對，最後不相等的一定要是兩對，而且這兩對互為相反。
python程式碼：

class Solution(object):
    def buddyStrings(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: bool
        """
        if len(A)!=len(B):
            return False
        if A==B and len(A)!=len(set(A)):
            return True
        dif=[(a,b) for a,b in zip(A,
 
B) if a!=b]
        return len(dif)==2 and dif[0]==dif[1][::-1]        

c++程式碼：

class Solution {
public:
    bool buddyStrings(string A, string B) {
        int len_A=A.size(),len_B=B.size();
        set<int>set_A(A.begin(),A.end());
        if(len_A!=len_B)
            return false;
        if(A==B && len_A!=set_A.size())
            return true;
        vector<vector<char>>dif;
        for(int i=0;i<len_A;i++)
        {
            if (A[i]!=B[i])
            {
                dif.push_back({A[i],B[i]});
            }
        }
        
        if (dif.size()!=2)
            return false;
        vector<char> dif1=dif[1];
        reverse(dif1.begin(),dif1.end());
        return dif[0]==dif1;
    }
};

總結：