标签：1.4 tom wrap div cte clu ons padding name
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
使用贪心的策略。使用两个变量：star——标记p字符串中*的位置， begin——标记s字符串中*匹配的的位置。首先推断是否同样。假设同样i++，j++。假设不同，则推断是否是*，记录begin 和 star值，j++。假设不是*，依据star和begin。又一次匹配，begin++。public boolean isMatch(String s, String p) { if ((s == null && p == null) || (s.length() == 0 && p.length() == 0)) { return true; } int lenS = s.length(); int lenP = p.length(); int star = -1; int begin = -1; int i = 0; int j = 0; while (i < lenS) { if (j < lenP && (s.charAt(i) == p.charAt(j) || p.charAt(j) == ‘?‘)) { i++; j++; } else if (j < lenP && p.charAt(j) == ‘*‘){ star = j; begin = i; j++; } else if (star != -1) { j = star + 1; i = begin + 1; begin++; } else { return false; } } while(j < lenP && p.charAt(j) == ‘*‘) { j++; } return j == lenP; }
标签：1.4 tom wrap div cte clu ons padding name
原文地址：http://www.cnblogs.com/jhcelue/p/6752156.html
