leetcode_c++：哈希： Isomorphic Strings(205)

最新推荐文章于 2020-04-25 09:39:52 发布

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本文介绍了一种用于判断两个字符串是否可以通过字符映射相互转换的高效算法。该算法通过一次遍历两个输入字符串实现O(n)的时间复杂度，并利用额外的字符映射数组来跟踪字符之间的对应关系。

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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.
// 判断两个字符是否可以直接映射

算法

o(n)

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        if(s.size()!=t.size())  return false;

        const int MAXCHAR = 256;
        char maps[MAXCHAR]={0}, mapt[MAXCHAR]={0};

        for(int i=0;i<s.size();i++){
            if(maps[s[i]]==0 && mapt[t[i]]==0){
                maps[s[i]]=t[i];
                mapt[t[i]]=s[i];
                continue;
            }

            if(maps[s[i]]==t[i] && mapt[t[i]]==s[i])
                continue;
            return false;
        }

        return true;

    }
};