Title

the problem link

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.

题目大意:判断是否存在这种对应关系，一一映射，注意是双向的。

题目和205差不多，不同的是那个是char-char，这个是char-string，所以用了两个map，其实也可以用一个map的，因为string大于一个字符，然后。。。。

Solution

//cpp:
class Solution {
public:
	bool wordPattern(string pattern, string str) {
		unordered_map<char, int> map1;
		unordered_map<string, int> map2;
		istringstream istrstream(str);
		string t;
		vector<string> v;
		while (istrstream >> t) {
			v.push_back(t);
		}
		if (pattern.length() != v.size())
			return false;
		int i = 0;
		for (i = 0; i < pattern.length(); i++) {
			if (i >= v.size() || map1[pattern[i]] != map2[v[i]]) {
				return false;
			}
			map1[pattern[i]] = i + 1;
			map2[v[i]] = i + 1;
		}
		return true;
	}
};

public boolean wordPattern(String pattern, String str) {
    String[] words = str.split(" ");
    if (words.length != pattern.length())
        return false;
    Map index = new HashMap();
    for (Integer i=0; i<words.length; ++i)
        if (index.put(pattern.charAt(i), i) != index.put(words[i], i))
            return false;
    return true;
}