290 Rules of words

Title Description

Answer key

At the beginning of my solution, many things were not taken into account , There are some special examples .

Special examples

Example 1

 Input ： pattern = “abba”, str = “dog dog dog dog”
 Output ： true

My initial code is based on whether it appears again hashmap Of key To judge , It's no use judging this value, So this situation is unrecognizable .

Example 2

 Input  pattern = “aaa“ str = “dog dog dog dog”
 Output ： false

At the beginning, I honestly forgot to judge these basic conditions, resulting in the length. I didn't want to wait and forgot to take it into account

Code

class Solution {
    
    public boolean wordPattern(String pattern, String s) {
    
        Map<Character, String> map = new HashMap<>();
        String[] arr = s.split(" ");
        int n = pattern.length();
        if (n != arr.length) {
    
            return false;
        }
        for (int i = 0; i < n; i++) {
    
            char ch = pattern.charAt(i);
            if (map.containsKey(ch)) {
    
                if (!Objects.equals(map.get(ch), arr[i])) {
    
                    return false;
                }
            } else {
    
                if (map.containsValue(arr[i])) {
    
                    return false;
                } else {
    
                    map.put(ch, arr[i]);
                }
            }
        }
        return true;
    }
}

Write your own code with leetcode What the boss wrote is still a lot worse , Here is leetcode The code written by the boss

class Solution {
    
    public boolean wordPattern(String pattern, String str) {
    
        String[] words = str.split(" ");
        // Words and characters map to each other , The quantity must be equal 
        if (words.length != pattern.length()) {
    
            return false;
        }
        Map<Object, Integer> map = new HashMap<>();
        for (Integer i = 0; i < words.length; i++) {
    
            /*  If key non-existent , Insert the success , return null; If key There is , Returns the previous corresponding value.  With pattern = "abba", str = "dog cat cat dog" For example ,  The first 1 Time ：map.put('a',0) return null,map.put("dog",0) return null, Both equal ;  The first 2 Time ：map.put('b',1) return null,map.put("cat",1) return null, Both equal ;  The first 3 Time ：map.put('b',2) return 1,map.put("cat",2) return 1, Both equal ;  The first 4 Time ：map.put('a',3) return 0,map.put("dog",3) return 0, Both equal ,  The result is  true.  With pattern = "abba", str = "dog cat cat fish" For example ,  The first 1 Time ：map.put('a',0) return null,map.put("dog",0) return null, Both equal ;  The first 2 Time ：map.put('b',1) return null,map.put("cat",1) return null, Both equal ;  The first 3 Time ：map.put('b',2) return 1,map.put("cat",2) return 1, Both equal ;  The first 4 Time ：map.put('a',3) return 0,map.put("fish",3) return null, The two are not equal ,  The result is  false. */
            if (map.put(pattern.charAt(i), i) != map.put(words[i], i)) {
    
                return false;
            }
        }
        return true;
    }
}

Why does it use Integer Rather than using int Well ?
If it is int Words , because If int The value of exceeds [-128,127] Words , Then this will recreate a new Integer object .
So it's directly used here Integer.