LeetCode 290. Word Pattern

原题链接在这里：https://leetcode.com/problems/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.

题解：

类似Isomorphic Strings.

通过建立HashMap 来存储 character 和 string 之间的对应来检查patter.

但这里要注意需要建立两个HashMap, 既要有character 到 string 的对应也要有string 到 character的对应。

e.g.  "aaaa", str = "dog cat cat dog" should return false. 但是若是没有 char 到 string 的对应就会返回true.

pattern = "abba", str = "dog dog dog dog" should return false. 但是若是没有 string 到 char的对应就会返回true.

Note: Character 类 也有一个equals(), 和 String的相同，使用起来也相同。

Time Complexity: O(pattern.length()). Space: O(pattern.length()).

AC Java:

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        if(pattern == null || pattern.length() == 0){
            return false;
        }
        if(str == null || str.length() == 0){
            return false;
        }
        String [] s = str.split("\\s+");
        if(pattern.length() != s.length){
            return false;
        }
        //hm1 contains string -> character pair
        HashMap<String,Character> hm1 = new HashMap<String,Character>();
        for(int i = 0; i<s.length; i++){
            if(!hm1.containsKey(s[i])){
                hm1.put(s[i],pattern.charAt(i));
            }else{
                if(!hm1.get(s[i]).equals(pattern.charAt(i))){
                    return false;
                }
            }
        }
        //hm2 contains character -> string pair
        HashMap<Character,String> hm2 = new HashMap<Character,String>();
        for(int i = 0; i<pattern.length(); i++){
            if(!hm2.containsKey(pattern.charAt(i))){
                hm2.put(pattern.charAt(i),s[i]);
            }else{
                if(!hm2.get(pattern.charAt(i)).equals(s[i])){
                    return false;
                }
            }
        }
        return true;
    }
}

AC C++:

class Solution {
public:
    bool wordPattern(string pattern, string s) {
        unordered_map<char, int> p2i;
        unordered_map<string, int> w2i;
        istringstream in(s);
        int i = 0;
        int n = pattern.size();
        for(string word; in >> word; i++){
            int a = p2i[pattern[i]];
            int b = w2i[word];
            if(i == n || p2i[pattern[i]] != w2i[word]){
                return false;
            }

            p2i[pattern[i]] = w2i[word] = i + 1;
        }

        return i == n;
    }
};

跟上Word Pattern II.