# 859. Buddy Strings 亲密字符串

@TOC

## # 题目描述

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.

## # 解题方法

### # 字典

1. 如果两个字符串长度不等，那么一定不满足条件
2. 如果两个字符串完全相等，如果其中存在至少两个相等字符，那么满足条件
3. 如果两个字符串长度相等且只有两个位置的字符不等，记录下这两个位置，如果这两个字符串的该两个位置字符是恰好错位的，那么满足条件。

``````class Solution:
def buddyStrings(self, A, B):
"""
:type A: str
:type B: str
:rtype: bool
"""
if len(A) != len(B):
return False
diff = 0
idxs = []
for i, a in enumerate(A):
if B[i] != a:
diff += 1
idxs.append(i)
counter = dict()
if diff == 0:
for a in A:
if a in counter and counter[a]:
return True
else:
counter[a] = True
if diff != 2:
return False
return A[idxs[0]] == B[idxs[1]] and A[idxs[1]] == B[idxs[0]]
``````

C++版本如下：

``````class Solution {
public:
bool buddyStrings(string A, string B) {
if (A.size() != B.size()) return false;
vector<int> ca(26);
vector<int> cb(26);
int N = A.size();
int diff = 0;
for (int i = 0; i < N; i++) {
if (A[i] != B[i] && diff++ > 2) return false;
ca[A[i] - 'a']++;
cb[B[i] - 'a']++;
}
for(int i = 0; i < 26; i++) {
if (diff == 0 && ca[i] > 1) return true;
if (ca[i] != cb[i]) return false;
}
return diff == 2;
}
};
``````

## # 日期

2018 年 7 月 4 日 —— 夏天挺热的，记得吃饭，防止低血糖 2018 年 11 月 30 日 —— 又到了周末