Bernard Wong
  • Software Engineer
  • Problems
    • Lowest Common Ancestor of a Binary Tree
    • Longest Substring Without Repeating Characters
    • Longest Palindromic Substring
    • Longest Common Prefix
    • Isomorphic Strings
    • Integer to Roman
    • Frog Jump
    • Find the Difference
    • Find k closest elements to a given value
    • Longest Common Subsequence
    • Binary Search Tree from Sorted Array
    • Balanced Binary Tree
    • Sort Using Two Stacks
    • O(1) Stack
    • k-th element to last of a LinkedList
    • Dedup LinkedList
    • Check Rotated String
    • Compress String by Character Count
    • Escape HTML whitespace
    • Check String Permutation
    • Unique String
    • Container With Most Water
    • 4 Sum
    • 3 Sum Closest
    • 3 Sum
    • 2 Sum
    • Maximum Subarray
    • Nested List Weight Sum
    • Palindrome Number
    • Pow(x, n)
    • Regular Expression Matching
    • Remove Nth Node From End of List
    • Reverse Integer
    • Roman to Integer
    • Rotate Array
    • Search a 2D Matrix
    • Shortest Word Distance
    • Two Sum III - Data structure design
    • Valid Parentheses
    • ZigZag Conversion
    • Quicksort
    • Add Two Numbers
    • Best Time to Buy and Sell Stock
    • Letter Combinations of a Phone Number
  • Data Structures
    • Heap
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  1. Problems

Regular Expression Matching

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).

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", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true
def isMatch(s: String, p: String): Boolean = {
  if (s.isEmpty) true
  else if (p.isEmpty) false
  else if (p.length >= 2 && p(1) == '*') {
    if (p.head == '.' || s.head == p.head) {
      isMatch(s.tail, p.substring(2)) || isMatch(s.tail, p)
    }
    else isMatch(s, p.substring(2))
  }
  else {
    if (p.head == '.' || s.head == p.head) {
      isMatch(s.tail, p.tail)
    }
    else false
  }
}

isMatch("aa", "a")
isMatch("aa", "aa")
isMatch("aaa", "aa")
isMatch("aa", "a*")
isMatch("aa", ".*")
isMatch("ab", ".*")
isMatch("aab", "c*a*b")

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Last updated 6 years ago