【题意】

给两个字符串s和p,判断s是否能用p进行匹配。

【题解】

dp[i][j]表示s的前i个是否能被p的前j个匹配。

首先可以分成3大类情况,我们先从简单的看起:

(1)s[i - 1] = p[j - 1],易得dp[i][j] = dp[i-1][j-1]

(2)p[i - 1] = '.',因为'.'可以匹配任何字符,所以dp[i][j] = dp[i-1][j-1]

(3)p[i - 1] = '*',这种情况就比较复杂了。

当p[j - 2] != s[i - 1] && p[j - 2] != '.'时,即当s = aba,p = abb*这种情况,因为 * 可以代表0个或多个,所以当p[j -2] != s[i - 1]时,往回倒退2个即dp[i][j] = dp[i][j - 2]

然后除去这种情况还有3种匹配情况,只要满足以下任意一种情况也是成立的

当 * 匹配多个时:dp[i][j] = dp[i - 1][j],即当s = abbb,p = ab*,因为b相等所以可以将s最后一个b删去,比较前面的abb与ab*从而实现 * 表示多个

当 * 匹配单个时:dp[i][j] = dp[i][j - 1],即当s = aba,p = aba*

当 * 匹配0个时:dp[i][j] = dp[i][j - 2],上面说过了

【代码】

 1 class Solution {
2 public:
3 bool dp[35][35];
4 bool isMatch(string s, string p) {
5 s = " " + s;
6 p = " " + p;
7 memset(dp , false, sizeof(dp));
8 dp[0][0] = true;
9 int m = s.size(), n = p.size();
10 for (int i = 1; i <= m; i++){
11 for (int j = 1; j <= n; j++){
12 if (s[i - 1] == p[j - 1])dp[i][j] = dp[i - 1][j - 1];
13 else if (p[j - 1] == '.')dp[i][j] = dp[i - 1][j - 1];
14 else if (p[j - 1] == '*'){
15 if (s[i - 1] != p[j - 2] && p[j - 2] != '.')
16 dp[i][j] = dp[i][j - 2];
17 else{
18 dp[i][j] = dp[i][j - 1] || dp[i][j - 2] || dp[i - 1][j];
19 }
20 }
21 }
22 }
23 return dp[m][n];
24 }
25 };

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