10. Regular Expression Matching

Problem

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

Solution

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
var isMatch = function(s, p) {
  var dp = Array(s.length + 1).fill(0).map(_ => Array(p.length + 1));
  return helper(dp, 0, 0, s, p);
};

var helper = function (dp, i, j, s, p) {
  var res = false;
  if (dp[i][j] !== undefined) return dp[i][j];
  if (j === p.length) {
    res = i === s.length;
  } else {
    if (i === s.length) {
      res = p[j + 1] === '*' && helper(dp, i, j + 2, s, p);
    } else if (s[i] === p[j] || p[j] === '.') {
      if (p[j + 1] === '*') {
        res = helper(dp, i + 1, j, s, p) || helper(dp, i, j + 2, s, p) || helper(dp, i + 1, j + 2, s, p);
      } else {
        res = helper(dp, i + 1, j + 1, s, p);
      }
    } else {
      res = p[j + 1] === '*' && helper(dp, i, j + 2, s, p);
    }
  }
  dp[i][j] = res;
  return res;
};

Explain:

动态规划，dp[i][j] 代表 s[i] 与 p[j] 是否可匹配。

Complexity:

• Time complexity : O(mn).
• Space complexity : O(mn).