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

這道題容易忽視兩個測試用例String A = "aa"; String B = "aa"和String A = "abab"; String B = "abab"，這兩類情況都應該返回true，發現了這點其實比較好通過。

此題大致分為三種情況：
1、A與B長度不等，這時候肯定返回false。
2、A與B完全相同，則檢視字串中是否有重複的字母，有的話就返回true，因為交換重複的字母不會改變字串，且滿足了題目要求。否則，返回false。
3、A中有且僅有兩個字母與B不同，且交換之後就與B相同了，這時候返回true。否則，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());
        }
    }
}