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 <= 200000 <= B.length <= 20000-
AandBconsist only of lowercase letters.
题目思路就是先用length来初步判断, 然后如果相等的话要看是否有重复的元素, 如果有那就可以, 否则不行, 比如aa, aa可以, ab, ab不行. 如果不等, 就要他们不同的地方正好两个, 并且交换顺序, 正好相等.
Code T: O(n) S; O(n) but if contains lower cases, then at most will be 26 , so you can explain it is O(1).
class Solution:
def buddyStrings(self, A, B):
la, lb = len(A), len(B)
if la != lb: return False
if A == B:
return any(val > 1 for val in collections.Counter(A).values())
ans = [None] * 2
for i in range(la):
if A[i] != B[i]:
if ans[1]:
return False
elif ans[0]:
ans[1]= A[i] + B[i]
else:
ans[0]= B[i] + A[i]
return ans[0] and ans[1] and ans[0] == ans[1]