10. Regular Expression Matching
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Given an input string (s
) and a pattern (p
), 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 lettersa-z
. -
p
could be empty and contains only lowercase lettersa-z
, and characters like.
or*
.
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 preceding 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 = "c*a*b"
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 = "mis*is*p*."
Output: false
难度系数
Hard
解法:动态规划
其中dp[i] [j]表示s[0,i)和p[0,j)是否match,注意左闭右开,状态转移方程如下
1. P[i][j] = P[i - 1][j - 1], if p[j - 1] != ‘*‘ && (s[i - 1] == p[j - 1] || p[j - 1] == ‘.‘);p和s当前位置字符相等或者p当前位置是‘.‘
2. P[i][j] = P[i][j - 2], if p[j - 1] == ‘*‘ 匹配0次,匹配0次:s=ab,p=aba*,将p的最后一个a去掉。
3. P[i][j] = P[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == ‘.‘), if p[j - 1] == ‘*‘ 匹配至少一次
具体代码如下
class Solution { public: bool isMatch(string s, string p){ int m = s.length(),n = p.length(); //其中dp[i][j]表示s[0,i)和p[0,j)是否match bool dp[m+1][n+1]; ? dp[0][0] = true; //初始化第0行,除了[0][0]全为false,毋庸置疑,因为空串p只能匹配空串,其他都无能匹配 for (int i = 1; i <= m; i++) dp[i][0] = false; //初始化第0列,只有X*能匹配空串,如果有*,它的真值一定和p[0][j-2]的相同(略过它之前的符号) for (int j = 1; j <= n; j++) dp[0][j] = j > 1 && ‘*‘ == p[j - 1] && dp[0][j - 2]; ? for (int i = 1; i <= m; i++){ for (int j = 1; j <= n; j++){ if (p[j - 1] == ‘*‘){ //dp[i][j - 2] 是匹配0次的情况,例子:s=ab,p=aba* //dp[i][j-1] 不匹配 例子:s=aba,p=aba* //s[i - 1] == p[j - 2] && dp[i - 1][j]是匹配至少一次,例子:s=abb,p=ab* //p[j - 2] == ‘.‘&& dp[i - 1][j]是匹配至少一次, 例子:s=abb,p=a.* dp[i][j] = dp[i][j - 2] || dp[i][j-1]||((s[i - 1] == p[j - 2] || p[j - 2] == ‘.‘) && dp[i - 1][j]); } else{ //p[j - 1] == ‘.‘ && dp[i - 1][j - 1] 例子:s=abb p=ab. //s[i - 1] == p[j - 1] && dp[i - 1][j - 1]例子:s=abb p=abb dp[i][j] = (p[j - 1] == ‘.‘ || s[i - 1] == p[j - 1]) && dp[i - 1][j - 1]; } } } return dp[m][n]; } };
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