leetcode-205. Isomorphic Strings

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

思路：两个数组分别存各自字符出现的次数，遍历判断，当某次个数不相等时返回false，否则为true

下面用的是hash表，效率会低一些

class Solution {
map<char,int> hashMap;
public:
bool isIsomorphic(string s, string t) {
//思路：将字符映射成对应数字，数字从1计数，存在hash表里，最后比较两个转换后的是否一样
int lengthS = s.size();
int lengthT = t.size();
if(lengthS != lengthT)
{
return false;
}
int globalIndex = 1;
//s转换
for(int i=0;i<lengthS;i++)
{
if(hashMap.find(s[i]) == hashMap.end())
{
hashMap.insert(make_pair(s[i],globalIndex));
s[i] = globalIndex;
globalIndex++;

}
else
{
s[i] = hashMap[s[i]];
}
}
//t转换,这个其实可以封装成函数，这冗余了先不改
hashMap.clear();
globalIndex = 1;
for(int i=0;i<lengthT;i++)
{
if(hashMap.find(t[i]) == hashMap.end())
{
hashMap.insert(make_pair(t[i],globalIndex));
t[i] = globalIndex;
globalIndex++;

}
else
{
t[i] = hashMap[t[i]];
}
}

//判断
for(int i=0;i<lengthS;i++)
{
if(s[i] != t[i])
{
return false;
}
}
return true;

}
};