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.
For example,
Given "egg"
, "add"
,
return true.
Given "foo"
, "bar"
,
return false.
Given "paper"
, "title"
,
return true.
Note:
You may assume both s and t have the same length.
方法:使用2个map即可。
class Solution { public: bool isIsomorphic(string s, string t) { map<char,char> ccmap,ccmap2; int i=s.length(),j=t.length(); if(i==j) { for(int m=0; m<i; m++) { map<char,char>::iterator iter=ccmap.find(s[m]); if(iter!=ccmap.end())//找到,此字符之前已经作为原始字符在替换中用过 { if((*iter).second!=t[m]) return false; } else ccmap[s[m]]=t[m]; } //通过map判断是否存在那种aa, ba这样的情况导致的多个字符映射到同一个字符的错误,如果存在,那么返回false int len=ccmap.size(),len2; //将ccmap中的key和value进行调换位置赋值给ccmap2,ccma自己不变 for(map<char,char>::iterator iter=ccmap.begin(); iter!=ccmap.end(); iter++) ccmap2[(*iter).second]=(*iter).first; len2=ccmap2.size(); if(len!=len2)//长度不等,那么说明ccmap中存在value相等的元素(如aa,ba: a-->b, a-->a,false) return false;// return true; } return false; } };
leetcode_Isomorphic Strings _easy
原文地址:http://blog.csdn.net/u013861066/article/details/46341173