[LeetCode]10. Regular Expression Matching

Description

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).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Example

isMatch(“aa”,”a”) → false
isMatch(“aa”,”aa”) → true
isMatch(“aaa”,”aa”) → false
isMatch(“aa”, “a*”) → true
isMatch(“aa”, “.*”) → true
isMatch(“ab”, “.*”) → true
isMatch(“aab”, “c*a*b”) → true

Discussion

这个问题可以采用动态规划。我们维护一个二位数组dp。dp[i+1][j+1]记录s[i]和p[j]之前的内容是否匹配。下面是dp的递推式：

dp初始都为0；
1. 若p和s当前字符匹配：dp[i + 1][j + 1] = dp[i][j]；
2. 若p当前为*，且p前一个字符a和s当前字符a匹配，则a*可以是0，1或n个：dp[i + 1][j + 1] = (dp[i + 1][j - 1] || dp[i + 1][j] || dp[i][j + 1])；
3. 若p当前为*，且p前一个字符a和s当前字符b不匹配，则a*只能是0个：dp[i + 1][j + 1] = dp[i + 1][j - 1]。

算法的时间复杂度为O(n^2)。

C++ Code

class Solution {public:    bool isMatch(string s, string p) {        //由dp[i+1][j+1]来记录s[i]和p[j]之前是否匹配        int dp[s.length() + 1][p.length() + 1];        memset(dp, 0, sizeof(dp));        dp[0][0] = 1;        //考虑a*b*c*开头可能为空的情况        for (int i = 1; i < p.length(); i++)        {            if (p[i] == '*' && dp[0][i-1])            {                dp[0][i+1] = 1;            }        }        for(int i = 0; i < s.length(); i++)        {            for(int j = 0; j < p.length(); j++)            {                //若p和s当前字符匹配                if(p[j] == s[i] || p[j] == '.')                {                    dp[i + 1][j + 1] = dp[i][j];                }                //若p当前为*                if(p[j] == '*')                {                    //若p前一个字符a和s当前字符a匹配，则a*可以是0，1或n个                    if(p[j - 1] == s[i] || p[j - 1] == '.')                    {                        dp[i + 1][j + 1] = (dp[i + 1][j - 1] || dp[i + 1][j] || dp[i][j + 1]);                    }                    //若p前一个字符a和s当前字符b不匹配，则a*只能是0个                    else                    {                        dp[i + 1][j + 1] = dp[i + 1][j - 1];                    }                }            }        }        return dp[s.length()][p.length()];    }};