205. Isomorphic Strings

问题描述

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.

题目链接：

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思路分析

判断连个字符串是否是同构的，两字符串长度相同。也就是把字母换成编码的话，两个字符串的编码是相同的

s-> 编码相同 <- t

因此我们用两个int的vector来表示两个string的编码，一开始都是置-1，因为0是代表着字母的。然后遍历string，将相应的编码置为第一次出现时的位置，然后在循环过程中对比，编码有没有出现错位，完成同构判断。

代码

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        vector<int> smap(256, -1);
        vector<int> tmap(256, -1);
        int n = s.length();
        for (int i = 0; i < n; i++){
            if (smap[s[i]] != tmap[t[i]])
                return false;
            smap[s[i]] = i;
            tmap[t[i]] = i;
        }
        return true;
    }
};

时间复杂度：$O(n)$
空间复杂度：$O(n)$

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反思

遇到了很大的阻力，没有想明白这道题目的思路。看过博客之后才明白了用数字代替string中字母的方式，然后就可以对字符相应位置的编码是否相同进行判断了。