Leetcode:44. Wildcard Matching

Description

Implement wildcard pattern matching with support for '?' and '*'.

Example

'?' 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

思路

• 思路一：递归，TLE
• 思路二：迭代回溯，两个指针
  • is：指向s的下一次迭代的位置，初始为同p的'*'对应位置，然后以后的每一次迭代加1
  • ip: 指向p的最近的一个'*'的位置

代码

• 递归超时。

bool isMatch(string s, string p) {
		int slen = s.size(), plen = p.size();
		string pattern;
		int j = 0;
		while (j < plen){
			if (j > 0 && p[j] == '*' && p[j - 1] == p[j]){
				j++;
				continue;
			}
			pattern += p[j++];
		}
		return helper(s, slen, 0, pattern, pattern.size(), 0);
	}

	bool helper(string &s, int slen, int i, string &p, int plen, int j){
		if (i == slen && j == plen) return true;
		if (j == plen) return false;
		if (i == slen) return p[j] == '*' ? helper(s, slen, i, p, plen, j + 1) : false;

		if (s[i] == p[j] || p[j] == '?')
			return helper(s, slen, i + 1, p, plen, j + 1);
		else if (p[j] == '*'){
			bool flag = false;
			while (j + 1 < plen && p[j + 1] == p[j]) j++;

			flag = helper(s, slen, i + 1, p, plen, j);
			if (!flag && j + 1 < plen && (p[j + 1] == s[i] || p[j + 1] == '?'))
				flag = helper(s, slen, i, p, plen, j + 1);

			return flag;
		}

		else return false;
	}

• 迭代回溯

class Solution {
public:
   bool isMatch(string s, string p) {
		int slen = s.size(), plen = p.size();
		
		int i = 0, j = 0, is = -1, ip = -1;
		while(i < slen){
		    if(p[j] == '*'){
		        is = i;
		        ip = j++;
		    }
		    else{
		        if(s[i] == p[j] || p[j] == '?'){
		            i++;
		            j++;
		        }
		        else if(ip == -1) return false;
		        else {
		            i = ++is;
		            j = ip + 1;
		        }
		    }
		}
		
		while(j < plen && p[j] == '*') j++;
		return j == plen;
	}
};