Wildcard Matching(Hard)
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", "cab") → false
解题思路1: Greedy
一个比较好的算法是贪心算法(greedy): whenever encounter "*" in p, keep record of the current position of * in p and the current index in s(mark). Try to match the stuff behind this * in p with s, if not matched, just s++ and then try to match again
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;
j++;
mark = i;
//这一步是关键,匹配s中当前字符与p中‘*’后面的字符,如果匹配,则在第一个if中处理,如果不匹配,则继续比较s中的下一个字符。
} else if (star != -1) {
j = star + 1;
i = ++mark;
} else {
return false;
}
}
//最后在此处处理多余的‘*’,因为多个‘*’和一个‘*’意义相同。
while (j < p.length() && p.charAt(j) == '*') {
++j;
}
return j == p.length();
}