[LeetCode] Word Pattern II
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:
- pattern =
"abab"
, str ="redblueredblue"
should return true. - pattern =
"aaaa"
, str ="asdasdasdasd"
should return true. - pattern =
"aabb"
, str ="xyzabcxzyabc"
should return false.
- pattern =
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 };
【推荐】还在用 ECharts 开发大屏?试试这款永久免费的开源 BI 工具!
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步