Wildcard Matching

描述

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

分析

跟上一题很类似。

主要是'*'的匹配问题。p每遇到一个'*'，就保留住当前'*'的坐标和s的坐标，然后s从前往后扫描，如果不成功，则s++，重新扫描。

递归版

Java

// Wildcard Matching
// 递归版，会超时，用于帮助理解题意
// 时间复杂度O(n!*m!)，空间复杂度O(n)
class Solution {
    public boolean isMatch(String s, String p) {
        return isMatch(s, 0, p, 0);
    }
    private boolean isMatch(String s, int i, String p, int j) {
        if (i == s.length() && j == p.length()) return true;
        if (i == s.length() || j == p.length()) return false;

        if (p.charAt(j) == '*') {
            while (j < p.length() && p.charAt(j) == '*') ++j;  //skip continuous '*'
            if (j == p.length()) return true;
            while (i < s.length() && !isMatch(s, i, p, j)) ++i;

            return i < s.length();
        }
        else if (p.charAt(j) == s.charAt(i) || p.charAt(j) == '?')
            return isMatch(s, ++i, p, ++j);
        else return false;
    }
}

C++

// Wildcard Matching
// 递归版，会超时，用于帮助理解题意
// 时间复杂度O(n!*m!)，空间复杂度O(n)
class Solution {
public:
    bool isMatch(const string& s, const string& p) {
        return isMatch(s.c_str(), p.c_str());
    }
private:
    bool isMatch(const char *s, const char *p) {
        if (*p == '\0' || *s == '\0') return *p == *s;
        else if (*p == '*') {
            while (*p == '*') ++p;  //skip continuous '*'
            if (*p == '\0') return true;
            while (*s != '\0' && !isMatch(s, p)) ++s;

            return *s != '\0';
        }
        else if (*p == *s || *p == '?') return isMatch(++s, ++p);
        else return false;
    }
};

迭代版

Java

// Wildcard Matching
// 迭代版，时间复杂度O(n*m)，空间复杂度O(1)
public class Solution {
    public boolean isMatch(String s, String p) {
        int i = 0, j = 0;
        int ii = -1, jj = -1;
        while (i < s.length()) {
            if (j < p.length() && p.charAt(j) == '*') {
                // skip continuous '*'
                while (j < p.length() && p.charAt(j) == '*') ++j;
                if (j == p.length()) return true;
                ii = i;
                jj = j;
            }
            if (j < p.length() && (p.charAt(j) == '?' || p.charAt(j) == s.charAt(i))) {
                ++i; ++j;
            } else {
                if (ii == -1) return false;
                ++ii;
                i = ii;
                j = jj;
            }
        }
        // skip continuous '*'
        while (j < p.length() && p.charAt(j) == '*') ++j;
        return i == s.length() && j == p.length();
    }
}

C++

// Wildcard Matching
// 迭代版，时间复杂度O(n*m)，空间复杂度O(1)
class Solution {
public:
    bool isMatch(const string& s, const string& p) {
        return isMatch(s.c_str(), p.c_str());
    }
private:
    bool isMatch(const char *s, const char *p) {
        bool star = false;
        const char *str, *ptr;
        for (str = s, ptr = p; *str != '\0'; str++, ptr++) {
            switch (*ptr) {
            case '?':
                break;
            case '*':
                star = true;
                s = str, p = ptr;
                while (*p == '*') p++;  //skip continuous '*'
                if (*p == '\0') return true;
                str = s - 1;
                ptr = p - 1;
                break;
            default:
                if (*str != *ptr) {
                    // 如果前面没有'*'，则匹配不成功
                    if (!star) return false;
                    s++;
                    str = s - 1;
                    ptr = p - 1;
                }
            }
        }
        while (*ptr == '*') ptr++;
        return (*ptr == '\0');
    }
};

相关题目

• Regular Expression Matching