somorphic Strings
Given two strings s and t, determine if they are isomorphic.
Two strings s and t 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
Constraints:
1 <= s.length <= 5 * 104
t.length == s.length
s and t consist of any valid ascii character.
思路一: 遍历字符串，不但要记录字符串还要记录字符串出现的次数，写的不太好
public boolean isIsomorphic(String s, String t) {
if (s.length() != t.length()) return false;
String c1 = count(s);
String c2 = count(t);
return c1.equals(c2);
}
private String count(String s) {
StringBuilder s1 = new StringBuilder();
char c = s.charAt(0);
int count = 1;
int x = 0;
Map<Character, Integer> map = new HashMap<>();
for (int i = 1; i < s.length(); i++) {
if (s.charAt(i) == c) {
count++;
} else {
if (map.containsKey(c)) {
x = map.get(c);
} else {
map.put(c, x);
}
s1.append((char)x).append(count);
c = s.charAt(i);
count = 1;
x = i;
}
}
if (count > 0) {
if (map.containsKey(c)) {
x = map.get(c);
}
s1.append((char)x).append(count);
}
return s1.toString();
}
思路二: 用数学上的双射，用两个 map 记录两个字符串映射关系，如果映射关系出现变化，map 也会一起更新，一旦出现不一致，说明字符不一样
public boolean isIsomorphic2(String s, String t) {
Map<Character, Character> s2t = new HashMap<>();
Map<Character, Character> t2s = new HashMap<>();
int len = s.length();
for (int i = 0; i < len; ++i) {
char x = s.charAt(i), y = t.charAt(i);
if ((s2t.containsKey(x) && s2t.get(x) != y) || (t2s.containsKey(y) && t2s.get(y) != x)) {
return false;
}
s2t.put(x, y);
t2s.put(y, x);
}
return true;
}