LeetCode 290. Word Pattern（单词模式）

最新推荐文章于 2024-04-28 21:49:44 发布

原创最新推荐文章于 2024-04-28 21:49:44 发布 · 651 阅读

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本文介绍了一种解决方案，用于判断给定的模式和字符串是否完全匹配。通过将模式和字符串映射为数字，并比较这些数字来实现。适用于LeetCode上的单词模式问题。

原题网址：https://leetcode.com/problems/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.

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.

方法：分别将pattern和str的单词映射到数字，然后比较数字是否一致。

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        Map<String, Integer> nums = new HashMap<>();
        int num = 1;
        String[] strs = str.split(" ");
        if (strs.length != pattern.length()) return false;
        for(int i=0; i<strs.length; i++) {
            if (!nums.containsKey(strs[i])) nums.put(strs[i], num++);
        }
        int[] strnum = new int[strs.length];
        for(int i=0; i<strs.length; i++) strnum[i] = nums.get(strs[i]);
        
        num = 1;
        int[] letters = new int[26];
        for(int i=0; i<pattern.length(); i++) {
            int j = pattern.charAt(i)-'a';
            if (letters[j] == 0) letters[j] = num ++;
        }
        int[] patternnum = new int[pattern.length()];
        for(int i=0; i<pattern.length(); i++) patternnum[i] = letters[pattern.charAt(i)-'a'];
        
        for(int i=0; i<patternnum.length; i++) {
            if (patternnum[i] != strnum[i]) return false;
        }
        
        return true;
    }
}