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 substring in str.
Examples:
pattern = "abab", str = "redblueredblue" should return true.
pattern = "aaaa", str = "asdasdasdasd" should return true.
pattern = "aabb", str = "xyzabcxzyabc" should return false.
Notes:
You may assume both pattern and str contains only lowercase letters.
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這次的word pattern不是之前那個已經給我們match好的pattern,而是需要我們自己去match並且check是否存在valid pattern。
所以要backtracking驗證。 code 如下:
class Solution {
public:
bool wordPatternMatch(string pattern, string str) {
unordered_set<string> strMp;
unordered_map<char, string> mp;
return match(pattern, str, 0, 0, strMp, mp);
}
private:
bool match(string& pattern, string& str, int strPos, int patPos, unordered_set<string>& strMp, unordered_map<char, string>& p2s ){
int m = str.size(), n = pattern.size();
if(strPos == m && patPos == n) return true;
if(n - patPos > m - strPos|| patPos == n) return false;// this condition reduce time!!
char cur = pattern[patPos];
// if cur char has matching string we try to verify it
if(p2s.find(cur) != p2s.end()){
string curStr = p2s[cur];
int len = curStr.size();
if(curStr != str.substr(strPos, len)) return false;
return match(pattern, str, strPos + len, patPos + 1, strMp, p2s);
}
for(int i = strPos; i < m; ++i){
string s = str.substr(strPos, i - strPos + 1);
if(strMp.find(s) != strMp.end()) continue;
strMp.insert(s);
p2s[cur] = s;
if(match(pattern, str, i+1, patPos+1, strMp, p2s)) return true;
strMp.erase(s);
p2s.erase(cur);
}
return false;
}
};