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Problem Description:
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:
"abab", str = "redblueredblue" should return true."aaaa", str = "asdasdasdasd" should return true."aabb", str = "xyzabcxzyabc" should return false. Notes:
You may assume both pattern and str contains only lowercase letters.
The problem becomes much more difficult after the spaces are removed. Now we need to determine which part matchs which part by ourselves. This post shares a very clear solution in Java, which is rewritten below in C++.
1 class Solution { 2 public: 3 bool wordPatternMatch(string pattern, string str) { 4 unordered_set<string> st; 5 unordered_map<char, string> mp; 6 return match(str, 0, pattern, 0, st, mp); 7 } 8 private: 9 bool match(string& str, int i, string& pat, int j, unordered_set<string>& st, unordered_map<char, string>& mp) { 10 int m = str.length(), n = pat.length(); 11 if (i == m && j == n) return true; 12 if (i == m || j == n) return false; 13 char c = pat[j]; 14 if (mp.find(c) != mp.end()) { 15 string s = mp[c]; 16 int l = s.length(); 17 if (s != str.substr(i, l)) return false; 18 return match(str, i + l, pat, j + 1, st, mp); 19 } 20 for (int k = i; k < m; k++) { 21 string s = str.substr(i, k - i + 1); 22 if (st.find(s) != st.end()) continue; 23 st.insert(s); 24 mp[c] = s; 25 if (match(str, k + 1, pat, j + 1, st, mp)) return true; 26 st.erase(s); 27 mp.erase(c); 28 } 29 return false; 30 } 31 };
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4870514.html
