leetcode-290-easy

Word Pattern

Given a pattern and a string s, find if s 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 s.

Example 1:

Input: pattern = "abba", s = "dog cat cat dog"
Output: true
Example 2:

Input: pattern = "abba", s = "dog cat cat fish"
Output: false
Example 3:

Input: pattern = "aaaa", s = "dog cat cat dog"
Output: false
Constraints:

1 <= pattern.length <= 300
pattern contains only lower-case English letters.
1 <= s.length <= 3000
s contains only lowercase English letters and spaces ' '.
s does not contain any leading or trailing spaces.
All the words in s are separated by a single space.

思路一: 双射，和前面一道字符串双射的问题一模一样，不过用了 split 函数，用了正则分割速度会慢很多，可以自己实现分割函数

public boolean wordPattern(String pattern, String s) {
    char[] pArr = pattern.toCharArray();
    String[] sArr = s.split("\\s");

    if (pArr.length != sArr.length) return false;

    Map<Character, String> m1 = new HashMap<>();
    Map<String, Character> m2 = new HashMap<>();
    for (int i = 0; i < pArr.length; i++) {
        char p = pArr[i];
        String si = sArr[i];

        if ((m1.containsKey(p) && m2.getOrDefault(si, Character.MIN_VALUE) != p) ||
                (m2.containsKey(si) && !m1.getOrDefault(p, "").equals(si))) {
            return false;
        }

        m1.put(p, si);
        m2.put(si, p);
    }

    return true;
}

原文地址：http://www.cnblogs.com/iyiluo/p/16825901.html