290. Word Pattern

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

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.

思路完全同205题：http://www.jianshu.com/p/6e03ef397be1

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(N)

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。

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

Time Complexity: O(N) Space Complexity: O(N)

Solution1 Code：

class Solution1 {
    public boolean wordPattern(String pattern, String str) {
        String[] words = str.split(" ");
        if (words.length != pattern.length()) return false;
        
        Map<Character, String> map_forw = new HashMap();
        Map<String, Character> map_bacw = new HashMap();
        
        for (int i = 0; i < pattern.length(); ++i) {
            if(!map_forw.containsKey(pattern.charAt(i)) && !map_bacw.containsKey(words[i])) {
                map_forw.put(pattern.charAt(i), words[i]);
                map_bacw.put(words[i], pattern.charAt(i));
            }
            else if(!map_bacw.containsKey(words[i]) || !map_bacw.get(words[i]).equals(pattern.charAt(i))) {
                    return false;
            }
        }
        return true;
    }
}

Solution2 Code：

class Solution {
    public boolean wordPattern(String pattern, String str) {
        String[] words = str.split(" ");
        if (words.length != pattern.length()) return false;
        
        Map<Character, String> map_forw = new HashMap();
        Set<String> back_set = new HashSet();
        
        for (int i = 0; i < pattern.length(); ++i) {
            System.out.println(i);
            if(!map_forw.containsKey(pattern.charAt(i)) && !back_set.contains(words[i])) {
                map_forw.put(pattern.charAt(i), words[i]);
                back_set.add(words[i]);
            }
            else if(!map_forw.containsKey(pattern.charAt(i)) || !map_forw.get(pattern.charAt(i)).equals(words[i])) {
                return false;
            }
        }
        return true;
    }
}

Solution3_a Code：

class Solution {
    public boolean wordPattern(String pattern, String str) {
        String[] words = str.split(" ");
        if (words.length != pattern.length()) return false;
        
        HashMap<Character, Integer> map1 = new HashMap<Character, Integer>();
        HashMap<String, Integer> map2 = new HashMap<String, Integer>();

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

Solution3_b Code：

class Solution {
    public boolean wordPattern(String pattern, String str) {
        String[] words = str.split(" ");
        if (words.length != pattern.length()) return false;
        
        HashMap<Character, Integer> map1 = new HashMap<Character, Integer>();
        HashMap<String, Integer> map2 = new HashMap<String, Integer>();

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