LeetCode 859. Buddy Strings

Given two strings s and goal, return true if you can swap two letters in s so the result is equal to goal, 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 s[i] and s[j].

• For example, swapping at indices 0 and 2 in "abcd" results in "cbad".

Example 1:

Input: s = "ab", goal = "ba"
Output: true
Explanation: You can swap s[0] = 'a' and s[1] = 'b' to get "ba", which is equal to goal.

Example 2:

Input: s = "ab", goal = "ab"
Output: false
Explanation: The only letters you can swap are s[0] = 'a' and s[1] = 'b', which results in "ba" != goal.

Example 3:

Input: s = "aa", goal = "aa"
Output: true
Explanation: You can swap s[0] = 'a' and s[1] = 'a' to get "aa", which is equal to goal.

Constraints:

• 1 <= s.length, goal.length <= 2 * 104
• s and goal consist of lowercase letters.

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题目要求两个string，能否通过只交换两个字符来变成另一个string。想来其实很简单，计算一下有几个字符不同，然后看看他们的下标是否能交换就行了。但是实际写起来踩了不少坑：

1. 在"aa"和"aa"这种情况，diffCount = 0，所以需要判断两个string是否相同，如果是相同的string那就要看这个string里有没有重复出现两或以上的字符 

2. 最开始写的时候，在循环里面就判断了if diffCount == 2直接看是否能swap就return，其实又可能到这儿都是符合的但是后面就不对应了，所以只能最后return了

3. char[diffCount]写成了char[i]……

class Solution {
    public boolean buddyStrings(String s, String goal) {
        int diffCount = 0;
        char[] diffS = new char[2];
        char[] diffGoal = new char[2];
        Set<Character> set = new HashSet<>();
        boolean repeat = false;
        if (s.length() != goal.length()) {
            return false;
        }
        for (int i = 0; i < s.length(); i++) {
            if (set.contains(s.charAt(i))) {
                repeat = true;
            }
            set.add(s.charAt(i));
            if (s.charAt(i) != goal.charAt(i)) {
                if (diffCount < 2) {
                    diffS[diffCount] = s.charAt(i);
                    diffGoal[diffCount] = goal.charAt(i);
                }
                diffCount++;
            }
        }
        if (diffCount == 0) {
            return repeat;
        } else if (diffCount == 2) {
            return diffS[0] == diffGoal[1] && diffS[1] == diffGoal[0];
        }
        return false;
    }
}