LEETCODE--[leetcode] Regular Expression Matching

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http://blog.csdn.net/liyi_/article/details/42736903

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)

Some examples:
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

solution: DP method. there are three main cases:

1. when *(p+1) == ‘*‘

      a. omitting ‘*‘ the preceding elements, considering the subsequent elements.

      b. if a doesn‘t work, pick ‘*‘ and preceding elements.

2. when *(p+1) != ‘*‘

3. else mismatch

bool isMatch(const char* s, const char* p) {
    if(s == NULL || p == NULL) 
　　　　　return false;
    if(*p == ‘\0‘ && *s == ‘\0‘) 
        return true;
    else {
        if( *(p+1) == ‘*‘ ) {
            if((*s != ‘\0‘ && *p == ‘.‘) || *s == *p) {
                //there may be zero or more preceding elements
                //we should take into account e.g: aaa-->a*a; aaa-->a*aa
                if(isMatch(s, p+2)) 
                    return true;
                return isMatch(++s, p);
            } else {
                return isMatch(s, p+2);
            }
        } else {
            if(*s == *p || (*s != ‘\0‘ && *p == ‘.‘)) {
                return isMatch(++s, ++p);
            }
        }
        return false;
    }
}

LEETCODE--[leetcode] Regular Expression Matching

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原文地址：http://www.cnblogs.com/michaelyili/p/4225714.html