leetcode-205-Isomorphic Strings

最新推荐文章于 2022-09-27 11:11:35 发布

别哔哔好伐

最新推荐文章于 2022-09-27 11:11:35 发布

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分类专栏： STL LeetCode 文章标签： leetcode

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本文介绍了一种算法，用于判断两个字符串是否具有映射关系。通过两种实现方式——使用map和数组，确保了字符一一对应且长度相等的条件下，能够正确判断两个字符串是否为同构字符串。

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.

Note:
You may assume both s and t have the same length.

两字符串有映射关系，即一一对应关系。

如egg,add.

e和a对应，g和d对应

刚开始用的map

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

后来想，用一个数组也一样，而且效率高了不少

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        int n =s.length();
        int m1[130] = {0},m2[130] = {0};
        for (int i = 0;i < n;i++) {
            m1[s[i]] = t[i];
            m2[t[i]] = s[i];
        }
        for (int i = 0;i < n;i++) {
            if (m1[s[i]] != t[i]) return false; 
            if (m2[t[i]] != s[i]) return false;
        }
        return true;
    }
};