【LeetCode】859. Buddy Strings 解题报告(Python)
标签(空格分隔): LeetCode
作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.me/
题目地址:https://leetcode.com/problems/buddy-strings/description/
题目描述:
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 <= 20000
- 0 <= B.length <= 20000
- A and B consist only of lowercase letters.
题目大意
当且仅当交换两个字符串中的两个字符的时候,看两个字符串能否完全相等。
解题方法
分析如下:
- 如果两个字符串长度不等,那么一定不满足条件
- 如果两个字符串完全相等,如果其中存在至少两个相等字符,那么满足条件
- 如果两个字符串长度相等且只有两个位置的字符不等,记录下这两个位置,如果这两个字符串的该两个位置字符是恰好错位的,那么满足条件。
代码如下:
class Solution:
def buddyStrings(self, A, B):
"""
:type A: str
:type B: str
:rtype: bool
"""
if len(A) != len(B):
return False
diff = 0
idxs = []
for i, a in enumerate(A):
if B[i] != a:
diff += 1
idxs.append(i)
counter = dict()
if diff == 0:
for a in A:
if a in counter and counter[a]:
return True
else:
counter[a] = True
if diff != 2:
return False
return A[idxs[0]] == B[idxs[1]] and A[idxs[1]] == B[idxs[0]]
日期
2018 年 7 月 4 日 ———— 夏天挺热的,记得吃饭,防止低血糖