题目描述
Given an input string (s) and a pattern §, 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
思路
dfs超时
动态规划，注意初始化p字符串前边多个为*的情况。
代码
class Solution {
public:
bool isMatch(string s, string p) {
int n = s.length();
int m = p.length();
vector<vector<int>> dp(n+1, vector<int>(m+1, 0));
dp[0][0] = 1;
int st = 0;
while(p[st] == '*') {
dp[0][st+1] = 1;
st++;
}
for (int i=1; i<=n; ++i) {
for (int j=1; j<=m; ++j) {
if (s[i-1] == p[j-1] || p[j-1] == '?') dp[i][j] |= dp[i-1][j-1];
if (p[j-1] == '*') {
dp[i][j] |= dp[i-1][j-1];
dp[i][j] |= dp[i-1][j];
dp[i][j] |= dp[i][j-1];
}
}
}
return dp[n][m];
}
};
