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.
Note:
You may assume both s and t have the same length.

For example,
Given "egg", "add", return true.
Given "foo", "bar", return false.
Given "paper", "title", return true.

思路完全同290题：//www.greatytc.com/p/4ffa25d11451

Solution1：two hashmap(forward mapping & backward mapping)

思路："a b c", "x y x": a -> x, b -> y, c -> z; x -> a, y -> b, z -> c
前向mapping，和 后向mapping 都要有。
前向避免：a -> x, a -> y的情况，后向避免a->x, b->x的情况。
(1) map_bacw.get(t.charAt(i)) == s.charAt(i)
(2) map_forw.get(s.charAt(i)) == t.charAt(i)
其实只check一个就行，Solution2中说明
Time Complexity: O(N) Space Complexity: O(字符种类)

Solution2：1 hashmap 1 hashset (forward mapping & "back" set)

(1)map_bacw.get(t.charAt(i)) == s.charAt(i)
(2)map_forw.get(s.charAt(i)) == t.charAt(i)
思路： 但其实不需要前向后向都map，因为a: x代表ax是一组的，只check a是否->x即可，因为如果a->x，x也肯定是和a一组的，因为只有在刚开始两者都不存在的时候在会建组(map)，就是说如果其中一个是有组的，就不会再建(不会和别的再组)。所以一个map即可，另一个反向的map只是为了表示是contain的，就是表示原来是出现过的有组了的。
所以可以两个hashmap，最后只检查一个条件(1)或(2)即可，或者直接将一个该为hashset。
Time Complexity: O(N) Space Complexity: O(字符种类)

Solution3：2 hashmap mapping element -> recent index

思路：相当于用index作为中间变量来表示他们是一组。
Note：实现时 如果把Integer改为int就不对了，因为map是<Object, Object>，如果传入int时会自动转成Integer。所以如果用int的话，两个int分别转为两个Integer，这样用!=判断的时候其实就当成了不同的变量return false (而不是判断实际值)。所以如果用int的话，需要用!...equals()来判断实际值，但不能null.equals()，所以还有判断取出的object是非null，如3_b所示，所以还是直接用Integer好
因为都是字符，共256中，3_c为将map替换为int数组做map

Time Complexity: O(N) Space Complexity: O(字符种类)

Solution1 Code：

public class Solution1 {
    public boolean isIsomorphic(String s, String t) {
        if(s == null || s.length() <= 1) return true;
        
        Map<Character, Character> map_forw = new HashMap();
        Map<Character, Character> map_bacw = new HashMap();
        
        for (int i = 0; i < s.length(); ++i) {
            if(!map_forw.containsKey(s.charAt(i)) && !map_bacw.containsKey(t.charAt(i))) {
                map_forw.put(s.charAt(i), t.charAt(i));
                map_bacw.put(t.charAt(i), s.charAt(i));
            }
            else if(!map_bacw.containsKey(t.charAt(i)) || !map_bacw.get(t.charAt(i)).equals(s.charAt(i))) {
                    return false;
            }
        }
        return true;
    }
}

Solution2 Code：

public class Solution2 {
    public boolean isIsomorphic(String s, String t) {
        if(s == null || s.length() <= 1) return true;
        
        Map<Character, Character> map_forw = new HashMap();
        Set<Character> back_set = new HashSet();
        
        for (int i = 0; i < s.length(); ++i) {
            if(!map_forw.containsKey(s.charAt(i)) && !back_set.contains(t.charAt(i))) {
                map_forw.put(s.charAt(i), t.charAt(i));
                back_set.add(t.charAt(i));
            }
            else if(!map_forw.containsKey(s.charAt(i)) || !map_forw.get(s.charAt(i)).equals(t.charAt(i))) {
                    return false;
            }
        }
        return true;
    }
}

Solution3_a Code：

public class Solution {
    public boolean isIsomorphic(String s1, String s2) {
        HashMap<Character, Integer> map1 = new HashMap<Character, Integer>();
        HashMap<Character, Integer> map2 = new HashMap<Character, Integer>();

        for (Integer i = 0; i < s1.length(); i++) {
            if (map1.get(s1.charAt(i)) != map2.get(s2.charAt(i))) return false;
            map1.put(s1.charAt(i), i);
            map2.put(s2.charAt(i), i);
        }
        return true;
    }
}

Solution3_b Code：

public class Solution {
    public boolean isIsomorphic(String s1, String s2) {
        HashMap<Character, Integer> map1 = new HashMap<Character, Integer>();
        HashMap<Character, Integer> map2 = new HashMap<Character, Integer>();

        for (int i = 0; i < s1.length(); i++) {
            if(!map1.containsKey(s1.charAt(i)) ^ !map2.containsKey(s2.charAt(i))) return false;
            
            if (map1.get(s1.charAt(i)) != null && !map1.get(s1.charAt(i)).equals(map2.get(s2.charAt(i)))) return false;
            map1.put(s1.charAt(i), i);
            map2.put(s2.charAt(i), i);
        }
        return true;
    }
}

Solution3_c Code：

public class Solution {
    public boolean isIsomorphic(String s1, String s2) {
        int[] map1 = new int[256];
        int[] map2 = new int[256];

        for (int i = 0; i < s1.length(); i++) {
            if(map1[s1.charAt(i)] != map2[s2.charAt(i)]) return false;
            map1[s1.charAt(i)] = map2[s2.charAt(i)] = i + 1; //avoid 0 when inited
        }
        return true;
    }
}