290. Word Pattern

290. Word Pattern           

Given a pattern and a stringstr, find ifstr follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter inpattern and anon-empty word instr.

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, andstr contains lowercase letters separated by a single space.

  

        这个题就是说判断是否字符串pattern中的每个元素与str中的每个元素存在一一对应的关系，即不同的KEY对应不同的VALUE，不太习惯用Map首先是用Vector实现的：

class Solution {public:bool wordPattern(string pattern, string str) {str = str + " ";string t;vector<string> s;for (int i = 0; i < str.size(); i++) {if (str[i] != ' ') t = t + str[i];else{s.push_back(t);t = "";}}if (pattern.size() != s.size()) return 0;for (int i = 0; i < pattern.size()-1; i++) {for (int j = i+1; j < pattern.size(); j++) {if (pattern[i] != pattern[j] && s[i] == s[j]) return 0;else if (pattern[i] == pattern[j] && s[i] != s[j]) return 0;}}return 1;}};

        就是将str存入字符数组s中，通过比较是否有一一对应关系得出结果。

        然后看了大神的代码：

bool wordPattern(string pattern, string str) {map<char, int> p2i;map<string, int> w2i;istringstream in(str);int i = 0, n = pattern.size();for (string word; in >> word; ++i) {if (i == n || p2i[pattern[i]] != w2i[word])return false;p2i[pattern[i]] = w2i[word] = i + 1;}return i == n;}

        确实很巧妙，通过两个Map来实现一一对应的关系。还是用Map简单。