标签：and length code pre string you div ash arrays
290. Word Pattern
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.
package leetcode.easy;
public class WordPattern {
@org.junit.Test
public void test() {
String pattern1 = "abba";
String str1 = "dog cat cat dog";
String pattern2 = "abba";
String str2 = "dog cat cat fish";
String pattern3 = "aaaa";
String str3 = "dog cat cat dog";
String pattern4 = "abba";
String str4 = "dog dog dog dog";
System.out.println(wordPattern(pattern1, str1));
System.out.println(wordPattern(pattern2, str2));
System.out.println(wordPattern(pattern3, str3));
System.out.println(wordPattern(pattern4, str4));
}
public boolean wordPattern(String pattern, String str) {
String[] arrays = str.split(" ");
if (arrays.length != pattern.length()) {
return false;
}
java.util.Map<String, Character> map = new java.util.HashMap<>();
for (int i = 0; i < pattern.length(); i++) {
if (map.containsKey(arrays[i]) && map.get(arrays[i]) != pattern.charAt(i)) {
return false;
} else if (!map.containsKey(arrays[i]) && map.containsValue(pattern.charAt(i))) {
return false;
} else {
map.put(arrays[i], pattern.charAt(i));
}
}
return true;
}
}
标签：and length code pre string you div ash arrays
原文地址：https://www.cnblogs.com/denggelin/p/11777309.html
