[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;
    }