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题解抄自“http://simpleandstupid.com/2014/10/26/wildcard-matching-leetcode-%E8%A7%A3%E9%A2%98%E7%AC%94%E8%AE%B0/”
“
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
Leetcode中最难的题之一，不过很多人onsite还是遇到了这道题，所以还是有必要掌握的。
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. Try to match the stuff behind this ‘*’ in p with s, if not matched, just s++ and then try to match again.
”
public class Solution { public boolean isMatch(String s, String p) { int star = -1, mark=-1, i=0, j=0; while(i<s.length()){ // if matched , both move on if(j<p.length() && (p.charAt(j)==‘?‘||p.charAt(j)==s.charAt(i))){ i++; j++; // if a star was found, put mark to i, star to j, move j to next }else if(j<p.length() && p.charAt(j)==‘*‘){ mark = i; star = j++; // if there was star#>=1, but nothing matched, 则继续比较s中的下一个字符 ：put j back to star‘s next position, move i to next, move mark to next(greedy, must check s 1by1) }else if(star!=-1){ j = star+1; i = ++mark; // no star, nothing matched }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() ; } }
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原文地址：http://www.cnblogs.com/jiajiaxingxing/p/4545392.html
