LeetCode Algorithms #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.

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.

思路：

将str按照空格切割成单词，然后同时遍历str切割后的数组与pattern中的单个字符，确保任意相同下标组合的pattern中的两字母与str切割后的两数组元素是否同时相等或不相等。

没有太多需要注意的，记得检验str切割后的数组元素个数与pattern的字符数是否相同以避免runtime error.

解：

class Solution {public:    std::vector<std::string> split(const std::string str, const std::string splitPattern){        std::vector<std::string> result;        if(!str.size())        {         return result;       }       size_t splitStart = 0;        size_t splitEnd = str.find_first_of(" ");        while (splitEnd != str.npos)        {            result.push_back(str.substr(splitStart, splitEnd - splitStart));           splitStart = splitEnd + 1;            splitEnd = str.find_first_of(splitPattern, splitStart);       }       if(!str.substr(splitStart).empty())       {           result.push_back(str.substr(splitStart));       }       return result;}    bool wordPattern(string pattern, string str)     {        std::vector<std::string> strChars = split(str, " ");        if(strChars.size() != pattern.size())        return false;        for(unsigned patternIndex = 0; patternIndex < pattern.size() - 1; patternIndex++)        {            for(unsigned compareIndex = patternIndex; compareIndex < pattern.size(); compareIndex++)            {                if(pattern[patternIndex] == pattern[compareIndex] && strChars[patternIndex] != strChars[compareIndex])                    return false;                if(pattern[patternIndex] != pattern[compareIndex] && strChars[patternIndex] == strChars[compareIndex])                    return false;            }        }    return true;    }};