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
通过率很低的一题,Java没指针写起来是不顺畅。
贪心的策略,能匹配就一直往后遍历,匹配不上了就看看前面有没有‘*‘来救救场,再从‘*‘后面接着试。
代码
public class WildcardMatching { public boolean isMatch(String s, String p) { int i = 0; int j = 0; int star = -1; int mark = -1; while (i < s.length()) { if (j < p.length() && (p.charAt(j) == ‘?‘ || p.charAt(j) == s.charAt(i))) { ++i; ++j; } else if (j < p.length() && p.charAt(j) == ‘*‘) { star = j++; mark = i; } else if (star != -1) { j = star + 1; i = ++mark; } else { return false; } } while (j < p.length() && p.charAt(j) == ‘*‘) { ++j; } return j == p.length(); } }
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原文:http://blog.csdn.net/perfect8886/article/details/22689147