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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
0and2in"abcd"results in"cbad".
Example 1:
<pre>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. </pre>
Example 2:
<pre>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. </pre>
Example 3:
<pre>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. </pre>
Constraints:
1 <= s.length, goal.length <= 2 * 10<sup>4</sup>sandgoalconsist of lowercase letters.
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题目大意：
给定两字符串，交换一次使得他们相等
解题思路：
三种情况： 长度不等，完全相等（若至少有一个重复，即满足题意），两次不同
解题步骤：
N/A
注意事项：
- 三种情况： 长度不等，完全相等，两次不同
Python代码：
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7def buddyStrings(self, s: str, goal: str) -> bool:
if len(s) != len(goal):
return False
if s == goal and len(set(s)) < len(goal): # any dups
return True
diff = [(a, b) for a, b in zip(s, goal) if a != b]
return True if len(diff) == 2 and diff[0] == diff[1][::-1] else False
算法分析：
时间复杂度为O(n)，空间复杂度O(1)
