LeetCode 10 Regular Expression Matching (正则匹配 记忆化搜索 推荐)

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Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

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 = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

题目链接：https://leetcode.com/problems/regular-expression-matching/

题目分析：考虑模式串p的长度，分为以下四种情况

1. len(p) == 0，判断是否len(s) == 0 

2. len(p) == 1，判断是否len(s) == 1且 (p[0]=s[0] 或 p[0]='.')

3. len(p) > 1 && p[1] != '*'，判断当前的p和s的首字母是否匹配

4. len(p) > 1 && p[1] == '*'，由于*可消掉前面的字符，此处需要分两种情况讨论 a)消掉前面的字符 b)匹配前面的字符一次

最后需要消掉通配符'*'

暴力DFS，64ms，时间击败21.6%

class Solution {
    
    public static boolean isFirstPosMatch(String s, String p) {
        return !s.isEmpty() && (s.charAt(0) == p.charAt(0) || p.charAt(0) == '.');
    }
    
    public boolean isMatch(String s, String p) {
        if (p.isEmpty()) {
            return s.isEmpty();
        }
        if (p.length() == 1) {
            return s.length() == 1 && isFirstPosMatch(s, p);
        }
        if (p.charAt(1) != '*') {
            return isFirstPosMatch(s, p) && isMatch(s.substring(1), p.substring(1));
        }
        while (isFirstPosMatch(s, p)) {
            if (isMatch(s, p.substring(2))) {
                return true;
            }
            s = s.substring(1);
        }
        return isMatch(s, p.substring(2));
    }
}

显然substring影响了运行效率，改写用下标记录位置，11ms，时间击败92.6%

class Solution {
    
    public static boolean isFirstPosMatch(String s, int spos, String p, int ppos) {
        return spos < s.length() && (s.charAt(spos) == p.charAt(ppos) || p.charAt(ppos) == '.');
    }
    
    public boolean isMatch(String s, String p) {
        return isMatchHelper(s, 0, p, 0);
    }

    public boolean isMatchHelper(String s, int spos, String p, int ppos) {
        if (ppos == p.length()) {
            return spos == s.length();
        }
        if (ppos == p.length() - 1) {
            return spos == s.length() - 1 && isFirstPosMatch(s, spos, p, ppos);
        }
        if (p.charAt(ppos + 1) != '*') {
            return isFirstPosMatch(s, spos, p, ppos) &&
                isMatchHelper(s, spos + 1, p, ppos + 1);
        }
        while (isFirstPosMatch(s, spos, p, ppos)) {
            if (isMatchHelper(s, spos, p, ppos + 2)) {
                return true;
            }
            spos++;
        }
        return isMatchHelper(s, spos, p, ppos + 2);
    }
}

不难发现求解过程中可能存在重复子问题，对答案记忆化一下，1ms，时间击败100%

class Solution {
    
    public static boolean isFirstPosMatch(String s, int spos, String p, int ppos) {
        return spos < s.length() && (s.charAt(spos) == p.charAt(ppos) || p.charAt(ppos) == '.');
    }
    
    public boolean isMatch(String s, String p) {
        int[][] match = new int[s.length() + 1][p.length() + 1];
        return isMatchHelper(s, 0, p, 0, match);
    }

    public boolean isMatchHelper(String s, int spos, String p, int ppos, int[][] match) {
        if (match[spos][ppos] != 0) {
            return match[spos][ppos] == 1 ? true : false;
        }
        if (ppos == p.length()) {
            match[spos][ppos] = (spos == s.length()) ? 1 : -1;
            return match[spos][ppos] == 1;
        }
        if (ppos == p.length() - 1) {
            match[spos][ppos] = (spos == s.length() - 1 && isFirstPosMatch(s, spos, p, ppos)) ? 1 : -1;
            return match[spos][ppos] == 1;
        }
        if (p.charAt(ppos + 1) != '*') {
            match[spos][ppos] = (isFirstPosMatch(s, spos, p, ppos) &&
                isMatchHelper(s, spos + 1, p, ppos + 1, match)) ? 1 : -1;
            return match[spos][ppos] == 1;
        }
        while (isFirstPosMatch(s, spos, p, ppos)) {
            match[spos][ppos] = (isMatchHelper(s, spos, p, ppos + 2, match)) ? 1 : -1;
            if (match[spos][ppos] == 1) {
                return true;
            }
            spos++;
        }
        match[spos][ppos] = (isMatchHelper(s, spos, p, ppos + 2, match)) ? 1 : -1;
        return match[spos][ppos] == 1;
    }
}