[LeetCode]--205. Isomorphic Strings（同构字符串）

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.

方法一：用一个转换函数，将我们现在的字符串弄成0123…替代的字符串，比如abcca为01220，最后比较转换出来的字符串是否相等。

public boolean isIsomorphic1(String s, String t) { if (s.length() != t.length()) return false; if (transferStr(s).equals(transferStr(t))) return true; return false; } public String transferStr(String s) { Map<String, Integer> map = new HashMap<String, Integer>(); String str = "", strTemp = ""; int temp = 0; for (int i = 0; i < s.length(); i++) { strTemp = s.charAt(i) + ""; if (!map.containsKey(strTemp)) { map.put(s.charAt(i) + "", temp++); } str = str + map.get(strTemp); } return str; }

很可惜，想法是好的，但是超时不能Accept。

方法二：建一个map保存映射关系, 同时用一个set保持被映射的char, 保证同一个char 不会被映射两次。

public boolean isIsomorphic2(String s, String t) { if (s == null || t == null) return false; if (s.length() != t.length()) return false; Map<Character, Character> map = new HashMap<Character, Character>(); Set<Character> set = new HashSet<Character>(); char c1, c2; for (int i = 0; i < s.length(); i++) { c1 = s.charAt(i); c2 = t.charAt(i); if (map.containsKey(c1)) { if (map.get(c1) != c2) return false; } else { if (set.contains(c2)) return false; else { map.put(c1, c2); set.add(c2); } } } return true; }