博文中会简要介绍Leetcode P0205题目分析及解题思路。
“Isomorphic Strings”是一道比较简单的题目，我们可以利用两个HashTable来建立字符串s和字符串t之间的一对一特异性的映射关系。所谓“特异性”表示s中的字符只能有唯一的t中的一个字符来替换，反之同理。
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.
以下是Java的题解代码实现。
class Solution {
public boolean isIsomorphic(String s, String t) {
int L1 = s.length(), L2 = t.length();
if (L1 != L2)
return false;
Map<Character, Character> s2tMap = new HashMap<>();
Map<Character, Character> t2sMap = new HashMap<>();
for (int idx=0; idx<L1; ++idx) {
char sch = s.charAt(idx), tch = t.charAt(idx);
if (!s2tMap.containsKey(sch)) {
if (t2sMap.getOrDefault(tch, sch) != sch)
return false;
t2sMap.put(tch, sch);
s2tMap.put(sch, tch);
}
else {
if (s2tMap.get(sch) != tch)
return false;
}
}
return true;
}
}
以下是C++的题解代码实现。
class Solution {
public:
bool isIsomorphic(string s, string t) {
int L1 = s.length(), L2 = t.length();
if (L1 != L2)
return false;
unordered_map<char, char> s2t, t2s;
for (int idx=0; idx<L1; ++idx) {
char tch = t[idx], sch = s[idx];
if (s2t.count(sch) == 0) {
if (t2s.count(tch) > 0 && t2s[tch] != sch)
return false;
s2t[sch] = tch;
t2s[tch] = sch;
}
else {
if (s2t[sch] != tch)
return false;
}
}
return true;
}
};