LeetCode #44 - Wildcard Matching

题目描述：

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

外卡匹配， “?”匹配任意一个字符，“*”匹配任意一串字符，包括空字符串。

与正则表达式匹配类似，还是采用动态规划，x[i][j]表示s的前i个字符和p的前j个字符匹配。如果p[j]!=’*‘，那么s[i]必须与p[j]匹配，即(s[i]==p[j]||p[j]=='?')&&x[i-1][j-1]==true；如果p[j]=='*'，那么p[j]可以匹配空字符串,即x[i][j-1]==true即可，或者匹配任意长度的字符串，也就是从x[0][j]到x[i-1][j]存在一个为真，那么x[i][j]为真，实际操作判断更为简单，只需要当x[k-1][j]==true,令x[k][j]=true即可，那么从x[k-1][j]开始到x[x.size()-1][j]都为真。

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