标签：href hide mil car bsp mon fir 长度 tput
Given an input string (s) and a pattern (p), 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).
Note:
s could be empty and contains only lowercase letters a-z.p could be empty and contains only lowercase letters a-z, and characters like ? or *.Example 1:
Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
Example 2:
Input:
s = "aa"
p = "*"
Output: true
Explanation: ‘*‘ matches any sequence.
Example 3:
Input:
s = "cb"
p = "?a"
Output: false
Explanation: ‘?‘ matches ‘c‘, but the second letter is ‘a‘, which does not match ‘b‘.
Example 4:
Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first ‘*‘ matches the empty sequence, while the second ‘*‘ matches the substring "dce".
Example 5:
Input:
s = "acdcb"
p = "a*c?b"
Output: false
通配符匹配，?代表单个特定字符，*代表若干长度的字符串
一看就很像这题，所以想都没想用了dp，但是比较慢
1 class Solution { 2 public: 3 bool isMatch(string s, string p) { 4 s = " " + s, p = " " + p; 5 int l1 = s.length(), l2 = p.length(); 6 vector<vector<bool>>dp(l1); 7 for (int i = 0; i < l1; i++)dp[i].resize(l2); 8 for (int i = 0; i < l1; i++) 9 for (int j = 0; j < l2; j++) 10 dp[i][j] = false; 11 dp[0][0] = true; 12 for (int i = 1; i < l2; i++) { 13 if (p[i] == ‘*‘) { 14 int j = 0; 15 for (; j < l1; j++) 16 if (dp[j][i - 1])break; 17 if (j < l1) 18 while (j < l1) { 19 dp[j][i] = true; 20 j++; 21 } 22 } 23 else if (p[i] == ‘?‘) { 24 for (int j = 0; j < l1 - 1; j++) 25 if (dp[j][i - 1]) 26 dp[j + 1][i] = true; 27 } 28 else { 29 for (int j = 1; j < l1; j++) 30 if (s[j] == p[i] && dp[j - 1][i - 1]) 31 dp[j][i] = true; 32 } 33 } 34 return dp[l1 - 1][l2 - 1]; 35 } 36 };
我看见有解法是贪心，貌似比我这个快得多，但是我今天不想再看了，以后再说
19.2.4 [LeetCode 44] Wildcard Matching
标签：href hide mil car bsp mon fir 长度 tput
原文地址：https://www.cnblogs.com/yalphait/p/10351754.html
