[lintcode]638.Strings Homomorphism

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.

注意事项：

You may assume both s and t have the same length.

样例：

Given s = "egg", t = "add", return true.

Given s = "foo", t = "bar", return false.

Given s = "paper", t = "title", return true.

解题思路：

一开始我一直钻到牛角尖里面，一直想着字母一一对应的替换。

最近在看算法，这道题应该算是一道很简单的哈希表。

其实就是要记录每一个字母出现的次数和它对应的位置。

以egg和add为例：

'g'和'd'出现了两次，就是[1,2,2]，[1,2,2]

'foo':[1,2,2];bar:[1,1,1]

然后我又再一次陷入了这个存数组的漩涡里面。。。我试图搞一个数组，按照index把出现次数存进去。。。

后来在网上搜索了一下，其实只要按照string的顺序（位置对应），读出每一个char的出现次数（次数对应）就可以以了。

class Solution {
public:
    /*
     * @param s: a string
     * @param t: a string
     * @return: true if the characters in s can be replaced to get t or false
     */
    bool isIsomorphic(string s, string t) {
        // write your code here
       map <char, int> word1,word2; //<key:char类型,value:int类型>
       for(int i = 0; i<s.size(); i++){
           word1[s[i]]++;
           word2[t[i]]++;
       }
       for(int i = 0; i< s.size(); i++){
           if(word1[s[i]]!=word2[t[i]]){
               return false;
           }
       }
       return true;
    }
};