Regular Expression Matching. 正則表達式匹配.
一、題目
Given an input string (s) and a pattern §, implement regular expression matching with support for ‘.’ and ‘*’
‘.’ Matches any single character.
‘*’ Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
Note:
- s could be empty and contains only lowercase letters a-z.
- p could be empty and contains only lowercase letters a-z, and characters like . or *.
Example 1:
Input:
s = “aa”
p = “a”
Output: false
Explanation: “a” does not match the entire string “aa”.
Example 2:
Input:
s = “aa”
p = “a*”
Output: true
Explanation: ‘*’ means zero or more of the precedeng element, ‘a’. Therefore, by repeating ‘a’ once, it becomes “aa”.
Example 3:
Input:
s = “ab”
p = “.*”
Output: true
Explanation: “." means "zero or more () of any character (.)”.
Example 4:
Input:
s = “aab”
p = “cab”
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches “aab”.
Example 5:
Input:
s = “mississippi”
p = “misisp*.”
Output: false
題目翻譯：
給定一個字符串 (s) 和一個字符模式 §。實現支持 ‘.’ 和 ‘*’ 的正則表達式匹配。
‘.’ 匹配任意單個字符
‘*’ 匹配零個或多個前面的那一個元素
二、解題方案
思路：
回溯法:（設s(i)表示串s從位置i開始的後綴。p(j)同理）
1.當s爲空串時，此時p爲空串，匹配成功。
2.當p爲空時，此時s不爲空，匹配失敗。
3.如果s和p的第一個字符匹配時，即s[0] == p[0] 或 p[0] == ‘.’時，要對除第一個字符外的s(1)和p(1)遞歸匹配。
4.當p[1] == ‘*’
5.其它情況，匹配失敗。
代碼實現：
class Solution {
public:
bool isMatch(string s, string p) {
if(s.empty()) return p.empty() || p.size() >= 2 && p[1] == '*' && dfs(s,p.substr(2));
if(p.empty()) return false;
if(s[0] == p[0] || p[0] == '.') {
bool r = dfs(s.substr(1),p.substr(1));
if(r) return true;
}
if(!(p.size() >= 2 && p[1] == '*')) return false;
bool r = dfs(s,p.substr(2));
if(r) return true;
for(int c=0;c<s.size();++c){
if(s[c] == p[0] || p[0] == '.') {
bool r = dfs(s.substr(c+1),p.substr(2));
if(r) return true;
}else break;
}
return false;
}
};