leetcode--Isomorphic Strings

最新推荐文章于 2025-12-21 17:04:13 发布

转载最新推荐文章于 2025-12-21 17:04:13 发布 · 200 阅读

文章标签：

#leetcode

LeetCode 专栏收录该内容

30 篇文章

订阅专栏

本文介绍了一种算法，用于判断两个字符串是否为同构字符串。通过使用HashMap和HashSet，确保所有出现过的字符都被正确映射且不会产生冲突。提供了三种不同的实现方式，包括使用Map进行直接比较的方法。

题目：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.

分析：建立HashMap来保存映射关系--建立HashSet来保存被映射

one：元素遍历，若s元素在Map里 s元素Map的键不是t元素false；若元素不在Map里在被映射的set里false，否则分别加入Map及set。

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s==null||t==null){return false;}
        if(s.length()!=t.length())return false;
        Map<Character,Character> map = new HashMap<Character,Character>();
        Set<Character> set = new HashSet<Character>();
        
        for(int i=0;i<s.length();i++){
            char c1 = s.charAt(i);
            char c2 = t.charAt(i);
            
            if(map.containsKey(c1)){
                if(map.get(c1)!=c2) return false;
            }else{
                if(set.contains(c2))return false;
                else{
                    map.put(c1,c2);
                    set.add(c2);
                }
            }
        }
        return true;
    }
}

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        Map m = new HashMap();
        for (Integer i=0; i<s.length(); ++i)
            if (m.put(s.charAt(i), i) != m.put(t.charAt(i)+"", i))
             return false;
        return true;
    }
}

public class Solution {
    
    public boolean isIsomorphic(String sString, String tString) {

        char[] s = sString.toCharArray();
        char[] t = tString.toCharArray();

        int length = s.length;
        if(length != t.length) return false;

        char[] sm = new char[256];
        char[] tm = new char[256];

        for(int i=0; i<length; i++){
            char sc = s[i];
            char tc = t[i];
            if(sm[sc] == 0 && tm[tc] == 0){
                sm[sc] = tc;
                tm[tc] = sc;
            }else{
                if(sm[sc] != tc || tm[tc] != sc){
                    return false;
                }
            }
        }
        return true;
    }
}

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        int m1[256] = {0}, m2[256] = {0}, n = s.size();
        for (int i = 0; i < n; ++i) {
            if (m1[s[i]] != m2[t[i]]) return false;
            m1[s[i]] = i + 1;
            m2[t[i]] = i + 1;
        }
        return true;
    }
};

public boolean isIsomorphic(String s1, String s2) {
        Map<Character, Integer> m1 = new HashMap<>();
        Map<Character, Integer> m2 = new HashMap<>();
    
        for(Integer i = 0; i < s1.length(); i++) {

            if(m1.put(s1.charAt(i), i) != m2.put(s2.charAt(i), i)) {
                return false;
            }
        }
        return true;
    }