205. Isomorphic Strings.jpg
LeetCode 205. Isomorphic Strings
Description
Given two strings s and t, determine if they are isomorphic.
Two strings are isomorphic if the characters in s can be replaced to get t.
All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.
Example 1:
Input: s = "egg", t = "add"
Output: true
Example 2:
Input: s = "foo", t = "bar"
Output: false
Example 3:
Input: s = "paper", t = "title"
Output: true
描述
给定两个字符串 s 和 t，判断它们是否是同构的。
如果 s 中的字符可以被替换得到 t ，那么这两个字符串是同构的。
所有出现的字符都必须用另一个字符替换，同时保留字符的顺序。两个字符不能映射到同一个字符上，但字符可以映射自己本身。
示例 1:
输入: s = "egg", t = "add"
输出: true
示例 2:
输入: s = "foo", t = "bar"
输出: false
示例 3:
输入: s = "paper", t = "title"
输出: true
思路
- 这道题我们使用一个字典，键为s中的字符，值为t中的字符，我们要求键和值要一一对应.
# -*- coding: utf-8 -*-
# @Author: 何睿
# @Create Date: 2019-01-21 11:56:34
# @Last Modified by: 何睿
# @Last Modified time: 2019-01-21 19:36:15
class Solution:
def isIsomorphic(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
# 字典，键为s中的字符，值为t中的字符
# 键和值必须一一对应
res = {}
for x, y in zip(s, t):
# 若键存在，则其值一定要和y相等
if x in res:
if not res.get(x) == y: return False
else:
# 若y已经作为其他键的值，直接返回False，因为这里要求键值一一对应
if y in res.values(): return False
else: res[x] = y
# 上面的条件都满足，返回True
return True
