一、 问题描述
Leecode第十题，题目为：
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
问题理解为：
给定字符串s和模式p，用’.’ 和 ''实现他们之间的正则表达匹配。
‘.’ 匹配任何字母
''匹配前一个或者更多个元素的0
匹配应该覆盖整个输入字符串(而不是部分)
注意：
1、s可以是空的，并且只包含小写字母a-z；
2、p可以是空的，只包含小写字母a-z或.或*。
例 1:
输入
s = “aa”
p = “a”
输出: false
注: “a” 不匹配整个字符串 “aa”.
例 2:
输入:
s = “aa”
p = “a*”
输出: true
注: '*'表示前置元素重复零或多次, ‘a’. 因此，通过重复“a”一次，它就变成了 “aa”.
例 3:
输入:
s = “ab”
p = “."
输出: true
注: ".” 表示“任何字符(.)的零或多个(*)”。
例 4:
输入:
s = “aab”
p = “cab”
输出: true
注: c 可以重复0次, a 可以重复1次. 因此它匹配"aab".
例 5:
输入:
s = “mississippi”
p = “misisp*.”
输出: false
二、解题思路
逐个字符匹配
三、实现代码
class Solution {
public:
bool isMatch(string s, string p) {
int x = s.length(), y = p.length();
vector<vector<bool> > dp(x + 1, vector<bool> (y + 1, false));
dp[0][0] = true;
for (int i = 0; i <= x; i++)
for (int j = 1; j <= y; j++)
if (p[j - 1] == '*')
dp[i][j] = dp[i][j - 2] || (i > 0 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') && dp[i - 1][j]);
else dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
return dp[x][y];
}
};
``