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English Version

题目描述

你有一个单词列表 words 和一个模式  pattern,你想知道 words 中的哪些单词与模式匹配。

如果存在字母的排列 p ,使得将模式中的每个字母 x 替换为 p(x) 之后,我们就得到了所需的单词,那么单词与模式是匹配的。

(回想一下,字母的排列是从字母到字母的双射:每个字母映射到另一个字母,没有两个字母映射到同一个字母。)

返回 words 中与给定模式匹配的单词列表。

你可以按任何顺序返回答案。

 

示例:

输入:words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
输出:["mee","aqq"]
解释:
"mee" 与模式匹配,因为存在排列 {a -> m, b -> e, ...}。
"ccc" 与模式不匹配,因为 {a -> c, b -> c, ...} 不是排列。
因为 a 和 b 映射到同一个字母。

 

提示:

  • 1 <= words.length <= 50
  • 1 <= pattern.length = words[i].length <= 20

解法

方法一:哈希表

Python3

class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        def match(s, t):
            m1, m2 = [0] * 128, [0] * 128
            for i, (a, b) in enumerate(zip(s, t), 1):
                if m1[ord(a)] != m2[ord(b)]:
                    return False
                m1[ord(a)] = m2[ord(b)] = i
            return True

        return [word for word in words if match(word, pattern)]

Java

class Solution {
    public List<String> findAndReplacePattern(String[] words, String pattern) {
        List<String> ans = new ArrayList<>();
        for (String word : words) {
            if (match(word, pattern)) {
                ans.add(word);
            }
        }
        return ans;
    }

    private boolean match(String s, String t) {
        int[] m1 = new int[128];
        int[] m2 = new int[128];
        for (int i = 0; i < s.length(); ++i) {
            char c1 = s.charAt(i);
            char c2 = t.charAt(i);
            if (m1[c1] != m2[c2]) {
                return false;
            }
            m1[c1] = i + 1;
            m2[c2] = i + 1;
        }
        return true;
    }
}

C++

class Solution {
public:
    vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
        vector<string> ans;
        auto match = [](string& s, string& t) {
            int m1[128] = {0};
            int m2[128] = {0};
            for (int i = 0; i < s.size(); ++i) {
                if (m1[s[i]] != m2[t[i]]) return 0;
                m1[s[i]] = i + 1;
                m2[t[i]] = i + 1;
            }
            return 1;
        };
        for (auto& word : words) if (match(word, pattern)) ans.emplace_back(word);
        return ans;
    }
};

Go

func findAndReplacePattern(words []string, pattern string) []string {
	match := func(s, t string) bool {
		m1, m2 := make([]int, 128), make([]int, 128)
		for i := 0; i < len(s); i++ {
			if m1[s[i]] != m2[t[i]] {
				return false
			}
			m1[s[i]] = i + 1
			m2[t[i]] = i + 1
		}
		return true
	}
	var ans []string
	for _, word := range words {
		if match(word, pattern) {
			ans = append(ans, word)
		}
	}
	return ans
}

TypeScript

function findAndReplacePattern(words: string[], pattern: string): string[] {
    return words.filter(word => {
        const map1 = new Map<string, number>();
        const map2 = new Map<string, number>();
        for (let i = 0; i < word.length; i++) {
            if (map1.get(word[i]) !== map2.get(pattern[i])) {
                return false;
            }
            map1.set(word[i], i);
            map2.set(pattern[i], i);
        }
        return true;
    });
}

Rust

use std::collections::HashMap;
impl Solution {
    pub fn find_and_replace_pattern(words: Vec<String>, pattern: String) -> Vec<String> {
        let pattern = pattern.as_bytes();
        let n = pattern.len();
        words
            .into_iter()
            .filter(|word| {
                let word = word.as_bytes();
                let mut map1 = HashMap::new();
                let mut map2 = HashMap::new();
                for i in 0..n {
                    if map1.get(&word[i]).unwrap_or(&n) != map2.get(&pattern[i]).unwrap_or(&n) {
                        return false;
                    }
                    map1.insert(word[i], i);
                    map2.insert(pattern[i], i);
                }
                true
            })
            .collect()
    }
}

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