44. Wildcard Matching

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

AC code:

class Solution {
public:
    bool isMatch(string s, string p) {
        int sp = 0;
        int pp = 0;
        int match = 0;
        int start = -1;
        while (sp < s.length()) {
            if (pp < p.length() && (s[sp] == p[pp] || p[pp] == '?')) {
                sp++;
                pp++;
            } else if (pp < p.length() && p[pp] == '*') {
                start = pp;
                match = sp;
                pp++;
            } else if (start != -1) {
                pp = start + 1;
                match++;
                sp = match;
            } else return false;
        }
        while (pp < p.length() && p[pp] == '*') {
            pp++;
        }
        return pp == p.length();
    }
};

Runtime: 24 ms, faster than 88.38% of C++ online submissions for Wildcard Matching.

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2021-05-05 23:23:08

44. 通配符匹配

给定一个字符串 (s) 和一个字符模式 (p) ，实现一个支持 '?' 和 '*' 的通配符匹配。

'?' 可以匹配任何单个字符。
'*' 可以匹配任意字符串（包括空字符串）。
两个字符串完全匹配才算匹配成功。

说明:

s 可能为空，且只包含从 a-z 的小写字母。
p 可能为空，且只包含从 a-z 的小写字母，以及字符 ? 和 *。
示例 1:

输入:
s = "aa"
p = "a"
输出: false
解释: "a" 无法匹配 "aa" 整个字符串。
示例 2:

输入:
s = "aa"
p = "*"
输出: true
解释: '*' 可以匹配任意字符串。
示例 3:

输入:
s = "cb"
p = "?a"
输出: false
解释: '?' 可以匹配 'c', 但第二个 'a' 无法匹配 'b'。
示例 4:

输入:
s = "adceb"
p = "*a*b"
输出: true
解释: 第一个 '*' 可以匹配空字符串, 第二个 '*' 可以匹配字符串 "dce".
示例 5:

输入:
s = "acdcb"
p = "a*c?b"
输出: false

思路：匹配串与模式串能否匹配主要有三种情况：

• if s[i] == p[j] || p[j] == '?'  =>  dp[i][j] = dp[i-1][j-1]
• if p[j] == '*'  =>  dp[i][j] = (dp[i-1][j] || dp[i][j-1])
• else dp[i][j] = false

 

其中，dp[i][j]表示的是s[0-i]与p[0-j]是否能够进行匹配。初始化的时候应该注意:

• dp[0][0] = true;
• dp[i][0] = false; i = 1, 2, ....
• if p[j] == '*'  =>  dp[0][j] = dp[0][j-1];  j = 1, 2, .....
• else dp[0][j] = false;

Code:

class Solution {
public:
    bool isMatch(string s, string p) {
        int len1 = s.length();
        int len2 = p.length();
        vector<vector<bool> > dp(len1+1, vector<bool>(len2+1, true));
        for (int i = 1; i <= len1; ++i) dp[i][0] = false;
        for (int i = 1; i <= len2; ++i) {
            if (p[i-1] == '*') dp[0][i] = dp[0][i-1];
            else dp[0][i] = false;
        }
        for (int i = 0; i < len1; ++i) {
            for (int j = 0; j < len2; ++j) {
                if (p[j] == '?' || s[i] == p[j]) {
                    dp[i+1][j+1] = dp[i][j];
                } else if (p[j] == '*') {
                    dp[i+1][j+1] = (dp[i][j+1] || dp[i+1][j]);
                } else {
                    dp[i+1][j+1] = false;
                }
            }
        }
        return dp[len1][len2];
    }
};