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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
思路：
这题其实比Regular Expression Matching那道题容易，判断p中"*"的情况更简单。
package dp; public class WildcardMatching { public boolean isMatch(String s, String p) { int slen = s.length(); int plen = p.length(); boolean[][] dp = new boolean[slen + 1][plen + 1]; dp[0][0] = true; for (int i = 1; i <= plen; ++i) dp[0][i] = p.charAt(i - 1) == ‘*‘ ? dp[0][i - 1] : false; for (int i = 1; i <= slen; ++i) dp[i][0] = false; for (int i = 1; i <= slen; ++i) { for (int j = 1; j <= plen; ++j) { if (p.charAt(j - 1) == ‘*‘) { dp[i][j] = dp[i - 1][j - 1] || dp[i - 1][j] || dp[i][j - 1]; } else { dp[i][j] = match(s.charAt(i - 1), p.charAt(j - 1)) && dp[i - 1][j - 1]; } } } return dp[slen][plen]; } private boolean match(char a, char b) { if (b == ‘?‘ || b == ‘*‘) return true; return a == b; } public static void main(String[] args) { // TODO Auto-generated method stub WildcardMatching w = new WildcardMatching(); System.out.println(w.isMatch("aa","a")); System.out.println(w.isMatch("aa","aa")); System.out.println(w.isMatch("aaa","aa")); System.out.println(w.isMatch("aa", "*")); System.out.println(w.isMatch("aa", "a*")); System.out.println(w.isMatch("ab", "?*")); System.out.println(w.isMatch("aab", "c*a*b")); } }
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原文地址：http://www.cnblogs.com/shuaiwhu/p/5075544.html
