10. Regular Expression Matching
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*
.
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 = "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
一道老题,然而想透却不太容易。首先问能不能匹配,给了两个序列,明显的是当前是否匹配和前部分是否匹配和当前字母有关联,是一道双序列DP题。该题的难点在于‘*’的处理。尤其是‘.*’的处理。这种题目如果在面试当中,一定要多问几个case彻底理解题意。比如Example3 Input: s = "ab", p = ".*", Output: true。可以看出‘.*’的组合,可以生成任意长度的字符去匹配s。而‘**’不能匹配任何s字符。
设置dp[i][j]对应前s的前i-1个字符是否匹配p的前j-1个字符
所以在求解过程中。针对当前字符是否为“*”做处理。转换状态为:
1. dp[i+1][j+1] = dp[i][j] if p[j] != '*' and (p[j] == '.' or s[i] == p[j]) (p[j]不是‘*’的情况)
2.dp[i+1][j+1] = dp[i+1][j-1] if p[j] == '*' and j > 0 (p[j]是‘*’,前面字符重复0次)
3.dp[i+1][j+1] = dp[i][j+1] if j > 0 and p[j-1] == '.' or s[i] == p[j-1] )(p[j]是‘*’,前面字符重复至少一次,既然前面字符重复至少一次,则把s[:i]去掉一个字符也可以和p[:j]匹配。
class Solution(object): def isMatch(self, s, p): """ :type s: str :type p: str :rtype: bool """ #把s, p为空的单独判断融合到主逻辑里面。之前是把s不为空,p为空 m = len(s) n = len(p) dp = [[False] * (n+1) for i in xrange(m+1)] dp[0][0] = True for i in xrange(n): # s="", p != "" if p[i] == '*' and (i > 0 and p[i-1] != '*') and dp[0][i-1] == True: dp[0][i+1] = True for i in xrange(m): for j in xrange(n): if p[j] != '*': if p[j] == '.' or s[i] == p[j]: dp[i+1][j+1] = dp[i][j] else: if j > 0 and dp[i+1][j-1] == True: # *匹配0个字符 dp[i+1][j+1] = True else: # *匹配1个以上字符 if j > 0 and p[j-1] == '.' or s[i] == p[j-1]: dp[i+1][j+1] = dp[i][j+1] return dp[m][n]
posted on 2018-09-24 12:30 Sheryl Wang 阅读(118) 评论(0) 编辑 收藏 举报