290-word-pattern.md

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 separated by a single space.

Swift代码

思路：双向校验。使用HashMap判断是否存在，不存在的话，存到两个HashMap里，存在是否和反向内容一致，不一致则返回false

func wordPattern(_ pattern: String, _ str: String) -> Bool {
        var strArr = str.components(separatedBy: " ");
        if pattern.count != strArr.count {
            return false
        }
        var keyValueDic = Dictionary<Character, String>()
        var keyValueDic2 = Dictionary<String, Character>()
        for (index,key) in pattern.enumerated() {
            let tempStr = keyValueDic[key]
            if tempStr == nil {
                if keyValueDic2[strArr[index]] != nil {
                    return false;
                }
                keyValueDic[key] = strArr[index]
                keyValueDic2[strArr[index]] = key
            }
            else {
                if tempStr == strArr[index] && keyValueDic2[tempStr!] == key {
                    continue
                }
                else {
                    return false
                }
            }
        }
        return true
    }

JAVA代码

思路同Swift

class Solution {
    public boolean wordPattern(String pattern, String str) {
         if(pattern.length()!=str.split(" ").length) return false;

        String[] array = str.split(" ");

        StringBuilder result = new StringBuilder();
        HashMap<Character, String> map = new HashMap<>();
        HashMap<String, String> mapWord = new HashMap<>();

        for(int i=0;i<pattern.length();i++){
            map.put(pattern.charAt(i), array[i]);
            mapWord.put(array[i], array[i]);
        }


        if(map.keySet().size()!=mapWord.keySet().size()){
            return false;
        }

        for (int i = 0; i < pattern.length(); i++) {
            if(i!=pattern.length()-1){
                result.append(map.get(pattern.charAt(i))).append(" ");
            }else {
                result.append(map.get(pattern.charAt(i)));
            }
        }
        return str.equals(result.toString());
    }
}