[LeetCode]Word Pattern

题目描述:（链接）

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:

1. pattern = "abba", str = "dog cat cat dog" should return true.
2. pattern = "abba", str = "dog cat cat fish" should return false.
3. pattern = "aaaa", str = "dog cat cat dog" should return false.
4. 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.

解题思路：

找到字符和字符串之间的映射关系就行了，用两个哈希表分别存放<字符，字符串>和<字符串, 字符>的映射关系，前提需要分解字符串(split函数)。

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        unordered_map<char, string> cache1;
        unordered_map<string, char> cache2;
        
        vector<string> splitStrs = split(str, ' ');
        
        if (splitStrs.size() != pattern.size()) {
            return false;
        }
        
        for (size_t i = 0; i < pattern.size(); ++i) {
            if (cache1.find(pattern[i]) == cache1.end() && cache2.find(splitStrs[i]) == cache2.end()) {
                cache1[pattern[i]] = splitStrs[i];
                cache2[splitStrs[i]] = pattern[i];
            } else if (cache1.find(pattern[i]) != cache1.end() && cache2.find(splitStrs[i]) != cache2.end()){
                if (cache1[pattern[i]] != splitStrs[i] && cache2[splitStrs[i]] != pattern[i]) {
                    return false;
                }
            } else {
                return false;
            }
        }
        
        return true;
    }
private:
    vector<string> split(const string& src, char sep) {
        vector<string> result;
        size_t lastIndex = 0;
        size_t index = 0;
        while ((index = src.find(sep, lastIndex)) != string::npos) {
            result.push_back(src.substr(lastIndex, index - lastIndex));
            lastIndex = index + 1;
        }
        
        string lastStr = src.substr(lastIndex);
        if (!lastStr.empty()) {
            result.push_back(lastStr);
        }
        
        return result;
    }
};