• [Leetcode][Python]44:Wildcard Matching


    # -*- coding: utf8 -*-
    '''
    __author__ = 'dabay.wang@gmail.com'

    44:Wildcard Matching
    https://oj.leetcode.com/problems/wildcard-matching/

    '?' 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

    ===Comments by Dabay===
    使用动态规划的做法.
    先想象在s和p的前面都加一个空格,以s为外循环,p为内循环。
    先初始化第一组数据,第一个“空”匹配“空”,所以第一个为True。然后看如果后面如果是*就继续设置为True。

    进入循环之后,判断根据p中的字符
    - 如果是?*以外的字符,就判断这个字符和s中对应的字符是否一致,而且s中到上一个字符和p中到上一次字符匹配,设True
    - 如果是?,如果s中到上一个字符和p中到上一个字符匹配,设True
    - 如果是*,满足下面的情况,就设True
    - (匹配0次)s中到这个字符和p到上两个字符匹配
    - (匹配1次)s中到这个字符和p到上一个字符匹配
    - (匹配n次)s中到上一个字符和p到这个字符匹配
    '''

    class Solution:
    # @param s, an input string
    # @param p, a pattern string
    # @return a boolean
    def isMatch(self, s, p):
    def quick_test(s, p):
    num_of_star = 0
    for x in p:
    if x == "*":
    num_of_star = num_of_star + 1
    return len(s) >= len(p)-num_of_star

    if quick_test(s, p) is False:
    return False
    default_row = [False] + [False for _ in p]
    current_row = list(default_row)
    current_row[0] = True
    for j in xrange(len(p)):
    if p[j] == "*":
    current_row[j+1] = True
    else:
    break
    previous_row = current_row
    for i in xrange(len(s)):
    current_row = list(default_row)
    for j in xrange(len(p)):
    if p[j] == "?":
    if previous_row[j]:
    current_row[j+1] = True
    elif p[j] == "*":
    if current_row[j] or previous_row[j] or previous_row[j+1]:
    current_row[j+1] = True
    else:
    if p[j] == s[i] and previous_row[j]:
    current_row[j+1] = True
    previous_row = current_row
    return previous_row[-1]


    def main():
    sol = Solution()
    s = "ac"
    p = "ab*"
    print sol.isMatch(s, p)


    if __name__ == "__main__":
    import time
    start = time.clock()
    main()
    print "%s sec" % (time.clock() - start)


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  • 原文地址:https://www.cnblogs.com/Dabay/p/4361229.html
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