leetcode 290： Word Pattern

问题描述：

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:

pattern = “abba”, str = “dog cat cat dog” should return true.

pattern = “abba”, str = “dog cat cat fish” should return false.

pattern = “aaaa”, str = “dog cat cat dog” should return false.

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：

首先把str给分割成单词数组，然后用笨方法，保存双向映射，这就是为什么代码中设置两个map。遍历数组，用map来查找和插入映射。

代码：

class Solution {
public:
bool wordPattern(string pattern, string str) {
//split into word list
vector<string> wordlist;
string::size_type pos;
str += " ";
int size = str.size();
for (int i = 0; i < size; i++)
{
pos = str.find(" ", i);  //find the space position
if (pos < size)
{
string word = str.substr(i, pos - i); //extract the word
if (word != "") //record the word
{
wordlist.push_back(word);
}
i = pos;
}
}
int w_size = wordlist.size();
int p_size = pattern.size();
if (w_size != p_size) return false;
map<char, string> c2s_mapped;  //map char to string
map<string, char> s2c_mapped;  //map string to char
for (int i = 0; i < p_size; i++)
{
char c_now = pattern[i];
string s_now = wordlist[i];
map<char, string>::iterator it_c2s = c2s_mapped.find(c_now);
if (it_c2s != c2s_mapped.end())  //found
{
if (it_c2s->second != s_now) return false;
}
else  //not found
{
map<string, char>::iterator it_s2c = s2c_mapped.find(s_now);
if (it_s2c != s2c_mapped.end()) return false;

c2s_mapped.insert(pair<char, string>(c_now, s_now));
s2c_mapped.insert(pair<string, char>(s_now, c_now));
}
}
return true;
}
};