290. Word Pattern

题目：

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:

1. pattern = "abba", str = "dog cat cat dog" should return true.
2. pattern = "abba", str = "dog cat cat fish" should return false.
3. pattern = "aaaa", str = "dog cat cat dog" should return false.
4. 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.

链接： http://leetcode.com/problems/word-pattern/

题解：

跟Isomophic Strings基本一样，使用两个hashmap保持pattern char和string的的一对一关系。  看到Discuss里也有使用一个HashMap的，非常巧妙，对put研究得很深。二刷要使用更好的方法。

Time Complexity- O(n)， Space Complexity - O(n)。

public class Solution {
    private Map<Character, String> patternToStr;
    private Map<String, Character> strToPattern;    
    
    public boolean wordPattern(String pattern, String str) {
        int len = pattern.length();
        String[] arrOfStr = str.split(" ");
        if(len != arrOfStr.length) {
            return false;
        }
        patternToStr = new HashMap<>();
        strToPattern = new HashMap<>();
        for(int i = 0; i < len; i++) {
            char c = pattern.charAt(i);
            if(!patternToStr.containsKey(c) && !strToPattern.containsKey(arrOfStr[i])) {
                patternToStr.put(c, arrOfStr[i]);
                strToPattern.put(arrOfStr[i], c);
            } else if(patternToStr.containsKey(c) && !arrOfStr[i].equals(patternToStr.get(c))) {
                return false;
            } else if(strToPattern.containsKey(arrOfStr[i]) && c != strToPattern.get(arrOfStr[i])) {
                return false;
            }
        }
        
        return true;
    }
}

 

二刷:

和一刷方法一样，也和Isomophic Strings一样

Java:

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        if (pattern == null || str == null) {
            return false;
        }
        String[] wordArr = str.split(" ");
        if (pattern.length() != wordArr.length) {
            return false;
        }
        Map<Character, String> pToWord = new HashMap<>();
        Map<String, Character> wordToP = new HashMap<>();
        for (int i = 0; i < pattern.length(); i++) {
            char p = pattern.charAt(i);
            String word = wordArr[i];
            if (!pToWord.containsKey(p)) {
                pToWord.put(p, word);
            } else if (!pToWord.get(p).equals(word)) {
                return false;
            }
            if (!wordToP.containsKey(word)) {
                wordToP.put(word, p);
            } else if (!wordToP.get(word).equals(p)) {
                return false;
            }
        }
        return true;
    }
}

 

三刷：

同上。

Java:

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

 

更好的写法来自Stefan Pochmann。

这里比较的是pattern里面的char， 和words里面的word上一次出现的位置是否相同。原理和Isomophic Strings一样。

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        if (pattern == null || str == null) {
            return false;
        }
        String[] words= str.split(" ");
        if (pattern.length() != words.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;
    }
}

 

Update:

重写了一下使用两个map

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        String[] words = str.split(" ");
        if (words.length != pattern.length()) return false;
        Map<Character, String> ps = new HashMap<>();
        Map<String, Character> sp = new HashMap<>();
        for (int i = 0; i < pattern.length(); i++) {
            char c = pattern.charAt(i);
            String word = words[i];
            if (!ps.containsKey(c)) ps.put(c, word);
            else if (!ps.get(c).equals(word)) return false;
            
            if (!sp.containsKey(word)) sp.put(word, c);
            else if (sp.get(word) != c) return false;
        }
        return true;
    }
}

 

利用Map.put，同时遍历pattern和words。 这里map.put()返回的是上一次保存的value，也就是上一次的index i

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

 

四刷:

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

 

 

 

Reference:

https://leetcode.com/discuss/62374/8-lines-simple-java

https://leetcode.com/discuss/62876/very-fast-3ms-java-solution-using-hashmap

https://docs.oracle.com/javase/6/docs/api/java/util/Map.html#put%28K,%20V%29