给定2个字符串，检查它们是否为同构字符串，C ++中的解决方案

如果第一个字符串中的一个字母可以被第二个字符串中的另一个字母替换，则说2个字符串是同构的。

Example 1:

string_1 = add
string_2 = egg

Output: True

Here “a” can be replaced by “e” and “d” can be replaced by “g”.

Example 2:

string_1 = foo
string_2 = bar

Output: False

Here “f” can be replaced by “b” and “o” can be replaced by,
“a”, and “o” again cannot be replaced by “r”.  Hence it will return false.

可以通过2个哈希图解决此问题。

我们将标记两个哈希图中元素所在的所有索引。
然后检查该索引处的两个哈希图是否相同。如果它们相同，我们将在下一个索引处进行检查。如果它们不相同，则表示我们已将该字母映射到另一个字母。因此退出循环。

在下面的输出中将给出有关实现的清晰思路。

C ++解决方案

#include<iostream>
#include<string>
#include<vector>

using namespace std;

bool check_isomorphic(string str_1, string str_2) 
{

	int hash_map_1[256]={0};
	int hash_map_2[256]={0};

	for(int i=0;i<str_1.length();i++)
	{
		cout<<"Step 1 i = "<<i <<" hash_map_1[str_1[i]] = "<< hash_map_1[str_1[i]]<<" hash_map_2[str_2[i]] = "<< hash_map_2[str_2[i]]<<endl;
		if(hash_map_1[str_1[i]]!=hash_map_2[str_2[i]])
		{ 
			return false;
		}
		
		hash_map_1[str_1[i]] = 1;
		hash_map_2[str_2[i]] = 1;
	} 

	return true;
}

int main()
{

	string str_1 = "add";
	string str_2 = "egg";

	cout<<"===========Test case 1==========="<<endl;
	cout<<"string 1 = "<<str_1 <<" string 2 = "<<str_2<<endl;

	if(check_isomorphic(str_1, str_2))
		cout<<"\nResult: The 2 strings are isomorphic"<<endl;
	else
		cout<<"\nResult: The 2 strings are NOT isomorphic"<<endl;


	str_1 = "foo";
	str_2 = "bar";

	cout<<"\n===========Test case 2==========="<<endl;
	cout<<"string 1 = "<<str_1 <<" string 2 = "<<str_2<<endl;

	if(check_isomorphic(str_1, str_2))
		cout<<"\nResult: The 2 strings are isomorphic"<<endl;
	else
		cout<<"\nResult: The 2 strings are NOT isomorphic"<<endl;



	return true;
}

输出：

===========Test case 1===========

string 1 = add string 2 = egg

Step 1 i = 0 hash_map_1[str_1[i]] = 0 hash_map_2[str_2[i]] = 0

Step 1 i = 1 hash_map_1[str_1[i]] = 0 hash_map_2[str_2[i]] = 0

Step 1 i = 2 hash_map_1[str_1[i]] = 1 hash_map_2[str_2[i]] = 1


Result: The 2 strings are isomorphic


===========Test case 2===========

string 1 = foo string 2 = bar

Step 1 i = 0 hash_map_1[str_1[i]] = 0 hash_map_2[str_2[i]] = 0

Step 1 i = 1 hash_map_1[str_1[i]] = 0 hash_map_2[str_2[i]] = 0

Step 1 i = 2 hash_map_1[str_1[i]] = 1 hash_map_2[str_2[i]] = 0


Result: The 2 strings are NOT isomorphic

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