290. Word Pattern

1.Question

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.

2.Code

codeA

class Solution {public:    bool wordPattern(string pattern, string str) {        unordered_map<char, int> map1;        unordered_map<string, int> map2;        int label = 1;        for(int i = 0; i < pattern.size(); i++)        {            if(str.empty()) return false;            char tempchar = pattern[i];            string tempstring;            int location = str.find(" ");            if(location == -1)            {                tempstring = str.substr(0, str.size());                str.erase(0, str.size() + 1);            }            else             {                tempstring = str.substr(0, location);                str.erase(0, location + 1);            }            if(map1.find(tempchar) != map1.end() || map2.find(tempstring) != map2.end())            {                if(map1[tempchar] != map2[tempstring]) return false;            }            else            {                map1[tempchar] = label;                map2[tempstring] = label;                label++;            }        }        return str.empty();    }};

codeB

class Solution {public:    bool wordPattern(string pattern, string str) {        unordered_map<char, int> map1;        unordered_map<string, int> map2;        istringstream stream(str);        int label = 1;        string word;        while(stream >> word)        {            char character = pattern[label - 1];            if(map1.find(character) != map1.end() || map2.find(word) != map2.end())            {                if(map1[character] != map2[word]) return false;            }            else            {                map1[character] = label;                map2[word] = label;            }            label++;        }        return label -1 == pattern.size();    }};

3.Note

a. 本题思想也比较简单，就是通过两个哈希表进行映射，让pattern 和str 有对应的映射。关键是，怎么把str 的单词分割出来。codeA 是最初想的，通过找分割标志符号来找单词的末位置，然后提出单词，并在str 中删掉，以便下次找单词。这种方法的好处是，分割标志符号不限定于" "。codeB 则利用了istringstream，但分隔标志符号只能是" "。具体用法看代码。

b. substr和erase的用法：substr(start, length).