205 Isomorphic Strings 同构字符串
Description:
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.
Example:
Example 1:
Input: s = "egg", t = "add"
Output: true
Example 2:
Input: s = "foo", t = "bar"
Output: false
Example 3:
Input: s = "paper", t = "title"
Output: true
Note:
You may assume both s and t have the same length.
题目描述:
给定两个字符串 s 和 t，判断它们是否是同构的。
如果 s 中的字符可以被替换得到 t ，那么这两个字符串是同构的。
所有出现的字符都必须用另一个字符替换，同时保留字符的顺序。两个字符不能映射到同一个字符上，但字符可以映射自己本身。
示例：
示例 1:
输入: s = "egg", t = "add"
输出: true
示例 2:
输入: s = "foo", t = "bar"
输出: false
示例 3:
输入: s = "paper", t = "title"
输出: true
说明:
你可以假设 s 和 t 具有相同的长度。
思路:
- 使用hashMap存放对应字符串
时间复杂度O(n), 空间复杂度O(n) - 使用 find()(Java为indexOf())
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
bool isIsomorphic(string s, string t)
{
for (int i = 0; i < s.size(); i++) if (s.find(s[i]) != t.find(t[i])) return false;
return true;
}
};
Java:
class Solution {
public boolean isIsomorphic(String s, String t) {
for (int i = 0; i < s.length(); i++) {
if (s.indexOf(s.charAt(i)) != t.indexOf(t.charAt(i))) return false;
}
return true;
}
}
Python:
class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
return all(s.find(s[i]) == t.find(t[i]) for i in range(len(s)))
