LintCode python 小白-简单题-638-Strings Homomorphism

题目：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.

样例
Given s = “egg”, t = “add”, return true.
Given s = “foo”, t = “bar”, return false.
Given s = “paper”, t = “title”, return true.

题目解析：给出两个字符串s和t，它们同构，（顺序相同，长度相同，结构相同）。

思路：使用字典进行映射关系的判断。主要是建立映射关系，和重复项关系判断。

1. 新建一个放s[i]为键，t[i]为值的字典dict和一个字符串string存放满足映射的t
2. 判断s[i]是否为dict的键,若是则代表dict中该键的值为t[i]
3. 若不是则先判断一个问题，t[i]是否是前面t值重复的，这时候就需要与字符串进行比较，判断是否是前面出现的t[0:t-1]，若存在，则代表s与t映射关系出现错误（如果存在映射关系，s[i]不在dict，代表t[i]的值不应该存在string），return False
4. 将s[i]、t[i]放在字典中，t[i]放在字符串string中
5. 跳转到2

代码如下：

class Solution:    # @param {string} s a string    # @param {string} t a string    # @return {boolean} true if the characters in s     # can be replaced to get t or false    def isIsomorphic(self, s, t):        # Write your code here        if len(s)<0 or len(t)<0 or len(s)!=len(t):            return False        dict={}             #建立一个空的字典        string=''           #建立一个空的字符串        for i in range(len(s)):            if s[i] not in dict.keys():   #判断s的第i-1个值是否为字典的健                if t[i] in string:                    return False                dict[s[i]]=t[i]                string=string+t[i]            else:                if dict[s[i]]!=t[i]:                    return False        return True