Isomorphic Strings
最后更新于:2022-04-01 22:56:21
## 一.题目描述
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.
## 二.题目分析
这道题主要是建立两个字符串中不同字符的对应关系,由于ASCII编码的字符只有256个,而且string中是不保存null字符的,因此,可以建立两个大小为256的字符数组来保存两个string中每一个字符在另外一个string中的对应的关系,然后遍历string中的每一个字符,如果相同位置的字符是互相对应的话,就处理下一个字符,如果不是互相对应的话,就在说明这两个string不是同等结构的。
也可以通过两个map来实现,但是map的查找过程时间复杂度为O(lgn),但是上面对于string中的每一个字符串都需要查找,因此,使用map的话,时间复杂度太高了。也可以使用hash表来做,也就是使用unordered_map来实现,但是由于ASCII编码的字符的个数是固定的而且个数比较少,使用数组完全可以很好地实现。
## 三.示例代码
~~~
class Solution
{
public:
bool isIsomorphic(string s, string t)
{
vector First(256, 0); // 创建一个含有256个0拷贝的vector
vector Second(256, 0);
for (size_t index = 0; index < s.size(); ++index)
{
unsigned char charOfFirst = s[index];
unsigned char charOfSecond = t[index];
unsigned char& First2Second = First[charOfFirst];
unsigned char& Second2First = Second[charOfSecond];
if (First2Second == 0 && Second2First == 0)
{
First2Second = charOfFirst;
Second2First = charOfSecond;
continue;
}
if (First2Second != 0 && Second2First != 0)
{
if (First2Second != charOfFirst && Second2First != charOfSecond)
return false;
continue;
}
return false;
}
return true;
}
};
~~~
示例结果:
![](https://docs.gechiui.com/gc-content/uploads/sites/kancloud/2016-01-05_568bb5eeedbdf.jpg)
![](https://docs.gechiui.com/gc-content/uploads/sites/kancloud/2016-01-05_568bb5ef0ce41.jpg)
## 四.小结
这道题就是寻找一个好的数据结构来保存两个string之间的字符的对应关系,根据这道题的假设,选择数组是一个比较的解决方案。
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