leetcode-10 Regular Expression Matching & leetcode-44 Wildcard Matching

leetcode-10 Regular Expression Matching 正则表达式匹配

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 *

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        # 边界条件，考虑 s 或 p 分别为空的情况
        if not p: return not s
        if not s and len(p) == 1: return False

        m, n = len(s) + 1, len(p) + 1
        dp = [[False for _ in range(n)] for _ in range(m)]
        # 初始状态
        dp[0][0] = True
        dp[0][1] = False

        for c in range(2, n):
            j = c - 1
            if p[j] == '*':
                dp[0][c] = dp[0][c - 2]
        
        for r in range(1,m):
            i = r - 1
            for c in range(1, n):
                j = c - 1
                if s[i] == p[j] or p[j] == '.':
                    dp[r][c] = dp[r - 1][c - 1]
                elif p[j] == '*':       # ‘*’前面的字符匹配s[i] 或者为'.'
                    if p[j - 1] == s[i] or p[j - 1] == '.':
                        dp[r][c] = dp[r - 1][c] or dp[r][c - 2]
                    else:                       # ‘*’匹配了0次前面的字符
                        dp[r][c] = dp[r][c - 2] 
                else:
                    dp[r][c] = False
        return dp[m - 1][n - 1]

class Solution {
public:  
     bool isMatch(string s, string p) {
        return regex_match(s, regex(p));
    }
};

leetcode-44 Wildcard Matching 通配符匹配

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 *.

方法1.

class Solution {
  public boolean isMatch(String s, String p) {
    int sLen = s.length(), pLen = p.length();
  
    if (p.equals(s) || p.equals("*")) return true;
    if (p.isEmpty() || s.isEmpty()) return false;

    // init all matrix except [0][0] element as False
    boolean[][] d = new boolean[pLen + 1][sLen + 1];
    d[0][0] = true;

    // DP compute
    for(int pIdx = 1; pIdx < pLen + 1; pIdx++) {
      if (p.charAt(pIdx - 1) == '*') {
        int sIdx = 1;
    
        while ((!d[pIdx - 1][sIdx - 1]) && (sIdx < sLen + 1)) sIdx++;
       
        d[pIdx][sIdx - 1] = d[pIdx - 1][sIdx - 1];
        while (sIdx < sLen + 1) d[pIdx][sIdx++] = true;
      }
      else if (p.charAt(pIdx - 1) == '?') {
        for(int sIdx = 1; sIdx < sLen + 1; sIdx++)
          d[pIdx][sIdx] = d[pIdx - 1][sIdx - 1];
      } 
      else {
        for(int sIdx = 1; sIdx < sLen + 1; sIdx++) {
          d[pIdx][sIdx] = d[pIdx - 1][sIdx - 1] &&
                  (p.charAt(pIdx - 1) == s.charAt(sIdx - 1));
        }
      }
    }
    return d[pLen][sLen];
  }
}

方法2.

算法：

我们使用两个指针：i 遍历输入字符串，j 遍历字符模式串。当 i < s_len：

如果字符模式仍有字符 j < p_len 且指针下的字符匹配 p[j] == s[i] 或 p[j] == '?'，则两个指针向前移动。
反之如果字符模式仍有字符 j < p_len 且 p[j] == '*'，则首先检查匹配 0 字符的情况，即只增加模式指针 j++。记下可能回溯的位置 jstar 和当前字符串的位置 istar。
反之如果出现不匹配的情况：
- 如果字符模式中没有星号，则返回 False。
- 如果有星号，则回溯：设置 j = jstar + 1 和 i = istar + 1，假设这次的星匹配多个字符。则可能的回溯为 istar = i。
如果字符模式的所有剩余字符都是星号，则返回 True。

例：s="abcdefghe" p="a*e"

class Solution {
public:
    bool isMatch(string s, string p) {
        int i = 0, j = 0, iStar = -1, jStar = -1, m = s.size(), n = p.size();
        while (i < m) {
            if (j < n && (s[i] == p[j] || p[j] == '?')) {
                ++i, ++j;
            } else if (j < n && p[j] == '*') {
                iStar = i;
                jStar = j++;
            } else if (iStar >= 0) {
                i = ++iStar;
                j = jStar + 1;
            } else return false;
        }
        while (j < n && p[j] == '*') ++j;//去除多余星号
        return j == n;
    }
};

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