Description
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:
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.
Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.
Solution
使用跟Isomorphic Strings相同的解法。感觉lower letters没有被用起来啊。。是不是哪里miss掉了？
Iterative, time O(n^2), space O(n)
重点：
- 利用string split方法分割word
- 利用HashMap的containsValue方法，注意是O(n)的复杂度哦。可以使用额外空间来避免该方法的调用。
class Solution {
public boolean wordPattern(String pattern, String str) {
String[] words = str.split(" ");
if (pattern.length() != words.length) {
return false;
}
Map<Character, String> map = new HashMap<>();
for (int i = 0; i < pattern.length(); ++i) {
char c = pattern.charAt(i);
String word = words[i];
if (!map.containsKey(c)) {
if (map.containsValue(word)) {
return false;
} else {
map.put(c, word);
}
} else if (!map.get(c).equals(word)) {
return false;
}
}
return true;
}
}
