044 Wildcard Matching

044 Wildcard Matching

这道题直观想法是用dp 复杂度 O(mn) 代码如下， 但是在用python的时候会超时

class Solution:
    # @param {string} s
    # @param {string} p
    # @return {boolean}
    def isMatch(self, s, p):
        m, n = len(p), len(s)
        dp = [False for i in range(0, n+1)]
        dp[0] = True
        for j in range(1,n+1):
            dp[j] = False
        for i in range(1, m+1):
            dpTemp = dp[:]
            if p[i-1] == '*':
                dp[0] = dpTemp[0]
            else:
                dp[0] = False
            isThereMatch = dpTemp[0]
            for j in range(1,n+1):
                isThereMatch = isThereMatch or dpTemp[j]
                if p[i-1] != '*':
                    if p[i-1] == '?':
                        dp[j] = dpTemp[j-1]
                    else:
                        dp[j] = (p[i-1]==s[j-1] and dpTemp[j-1])
                else:
                    dp[j] = isThereMatch
        return dp[n]

这道题有另外一个方法，比较巧妙， 具体复杂度我觉得是介于O(mn) 和 O(m+n)的范围内， 最后运行时间120ms 属于用python 里面最快的了， 方法很巧妙，临时很难想到。

class Solution:
    # @param s, an input string
    # @param p, a pattern string
    # @return a boolean
    def isMatch(self, s, p):
        s_cur, p_cur, starPos, lastMatch = 0, 0, -1, 0
        while s_cur<len(s):
            if p_cur< len(p) and (s[s_cur]==p[p_cur] or p[p_cur]=='?'):
                s_cur, p_cur = s_cur + 1, p_cur + 1
            elif p_cur < len(p) and p[p_cur]=='*':
                lastMatch = s_cur
                starPos = p_cur
                p_cur += 1
            elif  starPos != -1 :
                p_cur = starPos+1
                lastMatch = lastMatch+1
                s_cur = lastMatch
            else:
                return False
        while p_cur<len(p) and p[p_cur]=='*':
            p_cur = p_cur+1
             
        return p_cur == len(p)