(算法分析Week9)Wildcard Matching[Hard]

44.Wildcard Matching[Hard]

题目来源

Description

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).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false
           

Solution

这题和Regular Expression Matching差不多，区别在于’*’的用法，这里’*’能代替任何字符串，而不只是前一个字符的任意长度串。在这道题中，如果‘*’前面的字符没有匹配上，那么直接返回false了，根本不用继续往后搜寻。但是’*’可以匹配任意字符串，所以

if (p[i - 1] == '*') dp[0][i] = dp[0][i - 1];

略微修改一下代码即可。

Complexity analysis

O(M*N)

M = s.length()

N = p.length()

Code

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

Result