[LeetCode] 44. Wildcard Matching-优快云博客

本文链接：https://blog.youkuaiyun.com/Peng_maple/article/details/84801113

本文详细解析了通配符匹配算法的两种实现方法，包括动态规划和指针迭代。介绍了如何使用‘?’和‘*’进行模式匹配，以及在不同情况下如何判断一个字符串是否完全匹配给定的模式。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Wildcard Matching

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 = “ab”
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

解析

?可以匹配任意单个字符，*匹配任意长度的字符。判断p是否可以匹配s。

解法1：动态规划

使用动态规划求解：
使用一个二维数组dp[s.size+1][p.size+1]，dp[i][j]表示字符串s中0-i的字符能否匹配p中0-j的字符，并初始dp[0][0]=1;
当p[j]=s[i]或者p[j] =‘?’时，dp[i][j] = dp[i-1][j-1];
当p[j]=’*'时，可以匹配0个字符，dp[i][j] = dp[i-1][j]，可以匹配1个字符，dp[i][j] = dp[i-1][j-1]，所以
dp[i][j] = dp[i-1][j] || dp[i-1][j-1] ||…|| dp[i-1][0].
由于dp[i][j-1] = dp[i-1][j-1] || dp[i-1][j-2] ||…|| dp[i-1][0].
得到dp[i][j] = dp[i-1][j] || dp[i][j-1].
在这里插入图片描述

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

解法2

使用四个指针来记录位置，分别为
pcur：p字符串当前位置
scur：s字符串当前位置
pstar：最后一个*的位置
sstar：对应与pstar的s的位置
初始化pcur=scur=0，pstar=sstar=-1；
迭代过程如下：
当scur<s.size()时

如果s[scur]=p[pcur]或者p[pcur]=?，pcur++，scur++；
如果p[pcur]=‘*’，pstar=pcur，pcur++，sstar=scur；
如果pstar存在，pcur=pstar+1，sstar++，scur=sstar;

如果p[pcur]=’*’，pcur++；
如果pcur=p.size，返回true。

class Solution {
public:
    bool isMatch(string s, string p) {
        int sstar=-1;
        int pstar=-1;
        int pindex = 0,sindex=0;
        int size = s.size();
        while (sindex<size){
            if ((p[pindex]=='?')||(p[pindex]==s[sindex])){
                sindex ++;
                pindex ++;
            } 
            else if (p[pindex]=='*'){
                pstar = pindex; 
                pindex ++;
                sstar = sindex;
            } 
            else if (pstar != -1){ 
                pindex = pstar+1;
                sstar ++;
                sindex = sstar;
            } 
            else
                return false;
        }
        while (p[pindex]=='*'){pindex++;}
        return pindex==p.size();  
    }
};