Leetcode 205. Isomorphic Strings

Posted 2020-08-02 Deribs4

205. Isomorphic Strings

Total Accepted: 62899 Total Submissions: 206548 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.

 

思路：判断2个字符串是否同构。其实就是映射的问题，如果s和t的字符是一一映射（双射）的关系，就可以说s和t是同构的。

 

代码：

方法一：将s和t中的字符分别一一映射于整数集合，再对比映射以后的s和t是否相等。

 1 class Solution {
 2 public:
 3     string trans(string s){
 4         int i,m[256]={0};
 5         for(i=0;i<s.size();i++){
 6             if(!m[s[i]]){
 7                 m[s[i]]=i+1;
 8             }
 9             s[i]=m[s[i]];
10         }
11         return s;
12     }
13     bool isIsomorphic(string s, string t) {
14         if(trans(s)==trans(t))
15             return true;
16         return false;
17     }
18 };

 

方法二：
看是否能将s转换为t：

 1 class Solution {
 2 public:
 3     bool isIsomorphic(string s, string t) {
 4         char m[‘z‘+1]={0};
 5         bool b[‘z‘+1]={false};
 6         int i;
 7         for(i=0;i<s.size();i++){
 8             if(!m[s[i]]&&!b[t[i]]){
 9                 m[s[i]]=t[i];
10                 b[t[i]]=true;
11             }
12             if(m[s[i]]==t[i]){
13                 continue;
14             }
15             return false;
16         }
17         return true;
18     }
19 };

 

 

方法三：看当前字符最近一次出现的位置是否相等。

 1 class Solution {
 2 public:
 3     bool isIsomorphic(string s, string t) {
 4         int m1[256] = {0}, m2[256] = {0}, n = s.size();
 5         for (int i = 0; i < n; ++i) {
 6             if (m1[s[i]] != m2[t[i]]) return false;
 7             //因为初始值为0，所以要i+1
 8             m1[s[i]] = i + 1; //m1记录s[i]在s中最近一次出现的位置
 9             m2[t[i]] = i + 1;
10         }
11         return true;
12     }
13 };