【leetcode】【easy】290. Word Pattern

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本文详细解析了LeetCode上第290题“Word Pattern”的解题思路，通过使用map和set的数据结构，实现了一种高效的一一对应匹配算法，确保pattern与字符串str中单词的全匹配。

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 that may be separated by a single space.

题目链接：https://leetcode.com/problems/word-pattern/

解法：map

注意：特殊例子

"abba"
"dog dog dog dog"

结果是：false

因此可以看出pattern和句子是一一对应关系，这里对map中的value遍历。

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        unordered_map<char,string> record;
        vector<string> strs = split(str, ' ');
        if(pattern.length()!=strs.size()) return false;
        for(int i=0; i<strs.size();++i){
            if(record.find(pattern[i]) == record.end()) {
                if(hasValue(record, strs[i])) return false;
                record[pattern[i]] = strs[i];
            }
            else if(record[pattern[i]] != strs[i]) return false;
        }
        return true;
    }
    vector<string> split(string s,char token){
        istringstream iss(s);
        string word;
        vector<string> vs;
        while(getline(iss,word,token)){
            vs.push_back(word);
        }
        return vs;
    }
    bool hasValue(unordered_map<char,string> record, string value){
        for(unordered_map<char,string>::iterator i=record.begin();i!=record.end();++i){
            if(i->second==value) return true;
        }
        return false;
    }
};

改进：使用set来判断一一对应关系

降低时间复杂度

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        unordered_map<char,string> record;
        vector<string> strs = split(str, ' ');
        if(pattern.length()!=strs.size()) return false;
        unordered_set<string> values;
        for(int i=0; i<strs.size();++i){
            if(record.find(pattern[i]) == record.end()) {
                if(values.find(strs[i])!=values.end()) return false;
                record[pattern[i]] = strs[i];
                values.insert(strs[i]);
            }
            else if(record[pattern[i]] != strs[i]) return false;
        }
        return true;
    }
    vector<string> split(string s,char token){
        istringstream iss(s);
        string word;
        vector<string> vs;
        while(getline(iss,word,token)){
            vs.push_back(word);
        }
        return vs;
    }
};