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| 1 | +# [Wildcard Matching][title] |
| 2 | + |
| 3 | +## Description |
| 4 | + |
| 5 | +Implement wildcard pattern matching with support for `'?'` and `'*'`. |
| 6 | + |
| 7 | +``` |
| 8 | +'?' Matches any single character. |
| 9 | +'*' Matches any sequence of characters (including the empty sequence). |
| 10 | +
|
| 11 | +The matching should cover the entire input string (not partial). |
| 12 | +
|
| 13 | +The function prototype should be: |
| 14 | +bool isMatch(const char *s, const char *p) |
| 15 | +
|
| 16 | +Some examples: |
| 17 | +isMatch("aa","a") → false |
| 18 | +isMatch("aa","aa") → true |
| 19 | +isMatch("aaa","aa") → false |
| 20 | +isMatch("aa", "*") → true |
| 21 | +isMatch("aa", "a*") → true |
| 22 | +isMatch("ab", "?*") → true |
| 23 | +isMatch("aab", "c*a*b") → false |
| 24 | +``` |
| 25 | + |
| 26 | +**Tags:** String, Dynamic Programming, Backtracking, Greedy |
| 27 | + |
| 28 | + |
| 29 | +## 思路0 |
| 30 | + |
| 31 | +题意是让让你从判断`s`字符串是否通配符匹配于`p`，这道题和[Regular Expression Matching][010]很是相似，区别在于`*`，正则匹配的`*`不能单独存在，前面必须具有一个字符，其意义是表明前面的这个字符个数可以是任意个数，包括0个；而通配符的`*`是可以随意出现的，跟前面字符没有任何关系，其作用是可以表示任意字符串。在此我们可以利用*贪心算法*来解决这个问题，需要两个额外指针`p`和`match`来分别记录最后一个`*`的位置和`*`匹配到`s`字符串的位置，其贪心体现在如果遇到`*`，那就尽可能取匹配后方的内容，如果匹配失败，那就回到上一个遇到`*`的位置来继续贪心。 |
| 32 | + |
| 33 | +```java |
| 34 | +class Solution { |
| 35 | + public boolean isMatch(String s, String p) { |
| 36 | + if (p.length() == 0) return s.length() == 0; |
| 37 | + int si = 0, pi = 0, match = 0, star = -1; |
| 38 | + int sl = s.length(), pl = p.length(); |
| 39 | + char[] sc = s.toCharArray(), pc = p.toCharArray(); |
| 40 | + while (si < sl) { |
| 41 | + if (pi < pl && (pc[pi] == sc[si] || pc[pi] == '?')) { |
| 42 | + si++; |
| 43 | + pi++; |
| 44 | + } else if (pi < pl && pc[pi] == '*') { |
| 45 | + star = pi++; |
| 46 | + match = si; |
| 47 | + } else if (star != -1) { |
| 48 | + si = ++match; |
| 49 | + pi = star + 1; |
| 50 | + } else return false; |
| 51 | + } |
| 52 | + while (pi < pl && pc[pi] == '*') pi++; |
| 53 | + return pi == pl; |
| 54 | + } |
| 55 | +} |
| 56 | +``` |
| 57 | + |
| 58 | + |
| 59 | +## 思路1 |
| 60 | + |
| 61 | +另一种思路就是动态规划了，我们定义`dp[i][j]`的真假来表示`s[0..i)`是否匹配`p[0..j)`，其状态转移方程如下所示： |
| 62 | +* 如果`p[j - 1] != '*'`，`P[i][j] = P[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '?');` |
| 63 | +* 如果`p[j - 1] == '*'`，`P[i][j] = P[i][j - 1] || P[i - 1][j]` |
| 64 | + |
| 65 | +```java |
| 66 | +class Solution { |
| 67 | + public boolean isMatch(String s, String p) { |
| 68 | + if (p.length() == 0) return s.length() == 0; |
| 69 | + int sl = s.length(), pl = p.length(); |
| 70 | + boolean[][] dp = new boolean[sl + 1][pl + 1]; |
| 71 | + char[] sc = s.toCharArray(), pc = p.toCharArray(); |
| 72 | + dp[0][0] = true; |
| 73 | + for (int i = 1; i <= pl; ++i) { |
| 74 | + if (pc[i - 1] == '*') dp[0][i] = dp[0][i - 1]; |
| 75 | + } |
| 76 | + for (int i = 1; i <= sl; ++i) { |
| 77 | + for (int j = 1; j <= pl; ++j) { |
| 78 | + if (pc[j - 1] != '*') { |
| 79 | + dp[i][j] = dp[i - 1][j - 1] && (sc[i - 1] == pc[j - 1] || pc[j - 1] == '?'); |
| 80 | + } else { |
| 81 | + dp[i][j] = dp[i][j - 1] || dp[i - 1][j]; |
| 82 | + } |
| 83 | + } |
| 84 | + } |
| 85 | + return dp[sl][pl]; |
| 86 | + } |
| 87 | +} |
| 88 | +``` |
| 89 | + |
| 90 | + |
| 91 | +## 结语 |
| 92 | + |
| 93 | +如果你同我一样热爱数据结构、算法、LeetCode，可以关注我GitHub上的LeetCode题解：[awesome-java-leetcode][ajl] |
| 94 | + |
| 95 | + |
| 96 | + |
| 97 | +[010]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/010/README.md |
| 98 | +[title]: https://leetcode.com/problems/wildcard-matching |
| 99 | +[ajl]: https://github.com/Blankj/awesome-java-leetcode |
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