Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.

Examples:

pattern = "abba", str = "dog cat cat dog" should return true.
pattern = "abba", str = "dog cat cat fish" should return false.
pattern = "aaaa", str = "dog cat cat dog" should return false.
pattern = "abba", str = "dog dog dog dog" should return false.

Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

My original solution, using 2 HashMaps to store the relationship btw string s and pattern p:

class Solution {
    public boolean wordPattern(String pattern, String str) {
        String[] s = str.split(" ");
        char[] c = pattern.toCharArray();
        if(s.length!=c.length) return false;
        Map<Character, String> map = new HashMap<>();   
        Map<String, Character> pam = new HashMap<>();

        for(int i=0;i<s.length;i++){
            if(map.containsKey(c[i])){
                if(!map.get(c[i]).equals(s[i])) return false;
            }else{
                if(pam.containsKey(s[i])){
                    if(pam.get(s[i])!=c[i]) return false;
                }else {
                    pam.put(s[i],c[i]);
                    map.put(c[i],s[i]);
                }
            }
        }
        return true;
    }
}

But in fact to store the relatioship, we can store the index of strings, and use the previous index to find the realted character, which only needs 1 HashMap.

public boolean wordPattern(String pattern, String str) {
    String[] words = str.split(" ");
    if (words.length != pattern.length())
        return false;
    Map index = new HashMap();
    for (Integer i=0; i<words.length; ++i)
        if (index.put(pattern.charAt(i), i) != index.put(words[i], i))
            return false;
    return true;
}

Note:

You cannot simply compare 2 Integer objects, for “The JVM is caching Integer values. == only works for numbers between -128 and 127”, if the value exceed the boundary, the comparation will always return false.