0290. Word Pattern (E)

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.

Example 1:

Input: pattern = "abba", str = "dog cat cat dog"
Output: true

Example 2:

Input:pattern = "abba", str = "dog cat cat fish"
Output: false

Example 3:

Input: pattern = "aaaa", str = "dog cat cat dog"
Output: false

Example 4:

Input: pattern = "abba", str = "dog dog dog dog"
Output: false

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

题意

判断pattern中的字符是不是和str中的非空单词字符子串一一对应。

思路

用HashMap实现，注意必须要一对一，不存在多对一的情况。

代码实现

Java

class Solution {
    public boolean wordPattern(String pattern, String str) {
        Map<Character, String> hash = new HashMap<>();
        String[] s = str.split(" ");
        int i = 0;
        if (pattern.length() != s.length) {
            return false;
        }
        for (char c : pattern.toCharArray()) {
            if (!hash.containsKey(c)) {
                  // 必须一一对应，不能多对一
                if (hash.containsValue(s[i])) {
                    return false;
                }
                hash.put(c, s[i++]);
            } else if (!hash.get(c).equals(s[i])) {
                return false;
            } else {
                i++;
            }
        }
        return true;
    }
}

JavaScript

/**
 * @param {string} pattern
 * @param {string} str
 * @return {boolean}
 */
var wordPattern = function (pattern, str) {
  let map1 = new Map()
  let map2 = new Map()
  str = str.split(' ')
  if (str.length !== pattern.length) {
    return false
  }
  for (let i = 0; i < pattern.length; i++) {
    if (!map1.has(pattern[i]) && !map2.has(str[i])) {
      map1.set(pattern[i], str[i])
      map2.set(str[i], pattern[i])
    } else if (map1.get(pattern[i]) !== str[i]) {
      return false
    }
  }
  return true
}