标签：esc code sel eterm nsvalue desc rip lse char
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 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.
用hashmap，key是s[i]，value是t[i]
注意：如果map中已经存在s[i]这个key，不能再更新其value；由于一个value也只能对应一个key，如果map中已经存在s[i]对应的value，则无法添加新的映射到value的key
时间：O(N)，空间：O(N)
class Solution { public boolean isIsomorphic(String s, String t) { HashMap<Character, Character> map = new HashMap<>(); for(int i = 0; i < s.length(); i++) { if(map.containsKey(s.charAt(i)) && map.get(s.charAt(i)) == t.charAt(i)) continue; else if(!map.containsValue(t.charAt(i)) && !map.containsKey(s.charAt(i))) map.put(s.charAt(i), t.charAt(i)); else return false; } return true; } }
205. Isomorphic Strings - Easy
标签：esc code sel eterm nsvalue desc rip lse char
原文地址：https://www.cnblogs.com/fatttcat/p/10059982.html
