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Description
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
Input: A = "ab", B = "ba"
Output: true
Input: A = "ab", B = "ab"
Output: false
Input: A = "aa", B = "aa"
Output: true
Input: A = "aaaaaaabc", B = "aaaaaaacb"
Output: true
Input: A = "", B = "aa"
Output: false
Note
- 0 <= A.length <= 20000
- 0 <= B.length <= 20000
- A and B consist only of lowercase letters.
Solution
这道题容易忽视两个测试用例String A = "aa"; String B = "aa"和String A = "abab"; String B = "abab"，这两类情况都应该返回true，发现了这点其实比较好通过。
此题大致分为三种情况：
1、A与B长度不等，这时候肯定返回false。
2、A与B完全相同，则查看字符串中是否有重复的字母，有的话就返回true，因为交换重复的字母不会改变字符串，且满足了题目要求。否则，返回false。
3、A中有且仅有两个字母与B不同，且交换之后就与B相同了，这时候返回true。否则，返回false。
class Solution {
public boolean buddyStrings(String A, String B) {
if(A.length() != B.length())
return false;
HashSet<Character> set = new HashSet<>();
char[] diffA = new char[2];
char[] diffB = new char[2];
int index = 0;
for(int i = 0; i < A.length(); i++) {
set.add(A.charAt(i));
if(A.charAt(i) != B.charAt(i)){
if(index == 2) return false;
diffA[index] = A.charAt(i);
diffB[index] = B.charAt(i);
index++;
}
}
if(index == 2) {
return (diffA[0] != 0 && diffA[1] != 0 && diffA[0] == diffB[1] && diffB[0] == diffA[1]);
} else if(index == 1) {
return false;
} else {
return (A.equals(B) && set.size() < A.length());
}
}
}