44. Wildcard Matching.md

44. Wildcard Matching

• Difficulty: Hard.
• Related Topics: String, Dynamic Programming, Backtracking, Greedy.
• Similar Questions: Regular Expression Matching.

Problem

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 *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "*"
Output: true
Explanation: '*' matches any sequence.

Example 3:

Input:
s = "cb"
p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.

Example 4:

Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".

Example 5:

Input:
s = "acdcb"
p = "a*c?b"
Output: false

Solution

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

var test = function (s, p, sIndex, pIndex, dp) {
  if (dp[pIndex][sIndex] !== undefined) return dp[pIndex][sIndex];
  
  var sNow = s[sIndex];
  var pNow = p[pIndex];
  var res = false;

  if (pNow === undefined) return sNow === undefined;
  if (sNow === undefined) {
    for (var i = pIndex; i < p.length; i++) {
      if (p[i] !== '*') return false;
    }
    return true;
  }

  if (sNow === pNow || pNow === '?') {
    res = test(s, p, sIndex + 1, pIndex + 1, dp);
  } else if (pNow === '*') {
    res = test(s, p, sIndex, pIndex + 1, dp) || test(s, p, sIndex + 1, pIndex + 1, dp) || test(s, p, sIndex + 1, pIndex, dp);
  }

  dp[pIndex][sIndex] = res;

  return res;
};

Explain:

dp[i][j] 代表当正则第 i 位匹配字符串第 j 位时，是否 match.

Complexity:

• Time complexity : O(m*n).
• Space complexity : O(m*n).