[LeetCode September Challange]Day7-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.

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Solution

Time complexity : O(N) (size of pattern or str)
Space complexity : O(M) (unique strings in pattern and str)

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        vector<string> vec_str;
        istringstream iss(str);
        for (string s; iss >> s;)
            vec_str.push_back(s);
        if (vec_str.size() != pattern.length())
            return false;
        
        // use hash table to store first occurrence index.
        unordered_map<string, int> ht;
        for (int i=0; i<vec_str.size(); ++i) {
            string ptn_str = to_string(pattern[i]);
            if (ht.count(ptn_str) == 0)
                ht[ptn_str] = i;
            if (ht.count(vec_str[i]) == 0)
                ht[vec_str[i]] = i;
            if (ht[ptn_str] != ht[vec_str[i]])
                return false;
        }
        
        return true;
    }
};

首先將str切割成strings，裝入vector中。 判斷pattern長度與strings數量，若不相同，返回false。

使用一個hash table記錄第一次出現的pattern及string位置。
若沒有記錄過，就記錄上去。
若都有記錄過了，檢查第一次出現的位置是否相同，若不相同，則代表它們是不同的符號，返回false。

都檢查完畢，沒有問題，返回true。