<LeetCode OJ> 205. Isomorphic Strings

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205. Isomorphic Strings

Total Accepted: 46971 Total Submissions: 163994 Difficulty: Easy

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.

分析：

问题就出在有相同字符的地方，利用哈希map建立字符与位置的关系，如果字符重复的话，位置会变。

举例：

Given "egg", "add", return true.
1，先压入哈希map中得到每个字符及其对应的位置，011,011，

2，接着判断每个字符的位置是否一样011=011，所以true
Given "foo", "bar", return false.
1，压入哈希map得到位置序列，011,012
2，判断位置是否相等011！=012，所以false

class Solution {public:    bool isIsomorphic(string s, string t) {        if(s.size()!=t.size())            return false;        unordered_map<char,int> mapsing,mapting;        for(int i=0;i<s.size();i++)        {            mapsing[s[i]]=i;            mapting[t[i]]=i;         }        for(int i=0;i<s.size();i++)        {            if(mapsing[s[i]]!=mapting[t[i]])                return false;        }        return true;        }};

优化一下：更加耗时了，不解中......

class Solution {public:    bool isIsomorphic(string s, string t) {        if(s.size()!=t.size())            return false;        unordered_map<char,int> mapsing,mapting;        for(int i=0;i<s.size();i++)        {            if(mapsing[s[i]]!=mapting[t[i]])                return false;            mapsing[s[i]]=i+1;//这里之所以+1，是因为防止和初始值0相巧合，比如aa和ab            mapting[t[i]]=i+1;         }        return true;        }};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50463166

原作者博客：http://blog.csdn.net/ebowtang

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