LeetCode（44） Wildcard Matching

题目

Implement wildcard pattern matching with support for ‘?’ and ‘*’.

‘?’ Matches any single character.

‘*’ Matches any sequence of characters (including the empty sequence).

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”, “*”) → true

isMatch(“aa”, “a*”) → true

isMatch(“ab”, “?*”) → true

isMatch(“aab”, “c*a*b”) → false

分析

字符串匹配问题，有两种解决思路：

1. 采用递归实现，但是性能不好，TLE；
2. 贪心解决，参考网址

AC代码

class Solution {
public:
    //字符串模式匹配问题，s为匹配串，p为模式串
    bool isMatch(string s, string p) {
        return isMatch1(s.c_str(), p.c_str());
    }
    /*方法一：采用贪心的性质  AC*/
    bool isMatch1(const char *s, const char *p) {
        //? match one
        //* match 0,1,2,3..
        // aaaabc *c true
        const char* star = nullptr;
        const char* rs = nullptr;
        while (*s) {
            if (*s == *p || *p == '?') { //match
                s++; p++;
                continue;
            }
            if (*p == '*') {
                star = p; // record star
                p++; //match from next p
                rs = s; // record the position of s , star match 0
                continue;
            }
            if (star != nullptr) { //if have star in front then backtrace
                p = star + 1; //reset the position of p
                s = rs + 1;
                rs++; //star match 1,2,3,4,5....
                continue;
            }
            return false; //if not match return false
        }
        while (*p == '*') p++; //skip continue star
        return *p == '\0'; // successful match
    }
    /*方法二：利用字符串指针递归解决 性能不好 TLE*/
    bool isMatch2(const char *s, const char *p)
    {
        if (p == NULL || s == NULL)
            return false;
        else if (*p == '\0')
            return (*s == '\0');
        else
        {
            if (*p == '*')
            {
                while (*p == '*')
                    ++p;
                /*处理'*'可以匹配任意长度任意格式的字符串*/
                while (*s != '\0')
                {
                    if (isMatch2(s, p))
                        return true;
                    ++s;
                }//while
                return isMatch2(s, p);
            }
            else if((*s!='\0' && *p == '?') || (*s == *p)){
                return isMatch2(++s, ++p);
            }//else
            return false;
        }//else
    }
};

GitHub测试程序源码