29-Regular Expression Matching-leetcode

‘.’ Matches any single character.
‘*’ Matches zero or more of the preceding element.markdown

The matching should cover the entire input string (not partial).spa

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

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”) → truecode

题目难度:so hard!-参考别人思路
思路:
1.若下一个字符非星号,则必须彻底匹配模式串中字符或’.’,匹配完转化为断定下一轮子问题。
2..若下一个字符是星号,当前字符匹配可跳过星号进行匹配子问题,匹配成功返回,失败则以当前字符进行循环向前走再 转化为子问题求解,element

class Solution {
public:
    bool isMatch(string s, string p) {
        int ns=s.size(),np=p.size();
        if(np==0)return ns==0;
        int beg=0;
        int sn =0; 
        if(p[beg+1]!='*'){
            if(s[beg]==p[beg]||(p[beg]=='.'&&s[beg]!='\0'))
                return isMatch(string(s,beg+1,ns-1),string(p,beg+1,np-1));
            else return false;
        }
        else{
            while(s[sn]==p[beg]||(p[beg]=='.'&&s[sn]!='\0')){
                if(isMatch(string(s,sn,ns-sn),string(p,beg+2,np-2)))
                    return true;
                sn++;
            }
        }
        return isMatch(string(s,sn,ns-sn),string(p,beg+2,np-2));
    }
};
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