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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.
没做出来。
没有非常清楚的理解题目的意思,题目是说,两个字符串同构。我却认为是只要字符串s中哪些字符串一样,t中相同位置的字符串也要一样。
事实上约束更多,讲的是一种映射,就如同egg,e与a映射,g与d映射,那么字符串s中如果再出现e,字符串t中就必须是a。
所以,构建一个map,用上次见过的方法就行了。if(map.find(s[i])==map.end())****else****
不过需要有两次检查,检查s->t的映射t是否满足,还有t->s的映射s是否满足。
1 class Solution { 2 public: 3 bool isIsomorphic(string s, string t) { 4 map<char,char> model; 5 int slength=s.length(),tlength=t.length(); 6 if(slength!=tlength) return false; 7 for(int i=0;i<slength;i++){ 8 if(model.find(s[i])==model.end()) model[s[i]]=t[i]; 9 else if(model[s[i]]!=t[i]) return false; 10 } 11 model.clear(); 12 for(int i=0;i<slength;i++){ 13 if(model.find(t[i])==model.end()) model[t[i]]=s[i]; 14 else if(model[t[i]]!=s[i]) return false; 15 } 16 return true; 17 18 19 } 20 };
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原文地址:http://www.cnblogs.com/LUO77/p/5050356.html