44. Wildcard Matching

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).

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", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false

Solution:

Method 1: Two Pointers

We use two pointers.

One points to the string and one points to the pattern.

When we find a '?' in the pattern, we advance two pointers.

When we find a '*' in the pattern, we mark the current index of the string and the current index of the '*'. And start to do the match from mark and star + 1.

If we cannot find match, we go back to mark + 1 and star + 1 and try to match. (Every time we backtrack, we need to increase mark by 1.)

Worst case time complexity is O(mn).

The space complexity is O(1).



Code:

public class Solution {
    public boolean isMatch(String s, String p) {
        int i = 0;
        int j = 0;
        int mark = 0;
        int star = -1;
        while (i < s.length()) {
            if (j < p.length() && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '?')) {
                i++;
                j++;
            }
            else if (j < p.length() && p.charAt(j) == '*') {
                mark = i;
                star = j;
                j++;
            }
            else if (star != -1) {
                mark++;
                i = mark;
                j = star + 1;
            }
            else {
                return false;
            }
        }
        while (j < p.length() && p.charAt(j) == '*') {
            j++;
        }
        return j == p.length();
    }
}

Method 2: Dynamic Programming


Code: