【LeetCode】290. Word Pattern 解题报告（Python）

时间：2020-12-31 19:28:39

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目地址：https://leetcode.com/problems/word-pattern/#/description

题目描述

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.

题目大意

看模式和空格分割的字符串模式能不能构成一一映射。

解题方法

这个题第一感觉就是使用HashMap，看到Tags也是HashMap我就放心了。想法基本就是把字符和单词对应起来，看单词和字符能不能对应上，这样就知道模式是否匹配了。做的时候出现了一个小问题，就是pattern = "abba", str = "dog dog dog dog"这种，要判断是不是已经放进了value，之前不知道怎么办，今天学到了HashMap有个containsValue()方法，可以判断是否已经放进了value。

public class Solution {

    public boolean wordPattern(String pattern, String str) {

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

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

        if(words.length != pattern.length()){

            return false;

        }

        for(int i = 0; i < words.length; i++){

            String word = words[i];

            char temp = pattern.charAt(i);

            if(map.containsKey(temp)){

                if(!map.get(temp).equals(word)){

                    return false;

                }

            }else{

                if(map.containsValue(word)){

                    return false;

                }

                map.put(temp, word);

            }

        }

        return true;

    }

}

二刷，使用Python解法，这个题和205. Isomorphic Strings一模一样，所以很快就写出来这个一一映射的代码：

class Solution(object):

    def wordPattern(self, pattern, str):

        """

        :type pattern: str

        :type str: str

        :rtype: bool

        """

        strs = str.split()

        if len(pattern) != len(strs):

            return False

        d = dict()

        for i, p in enumerate(pattern):

            if p not in d:

                d[p] = strs[i]

            else:

                if d[p] != strs[i]:

                    return False

        d = dict()

        for i, p in enumerate(strs):

            if p not in d:

                d[p] = pattern[i]

            else:

                if d[p] != pattern[i]:

                    return False

        return True

日期

2017 年 5 月 19 日
2018 年 11 月 24 日 —— 周六快乐

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