LeetCode 44. Wildcard Matching

原题链接在这里：https://leetcode.com/problems/wildcard-matching/

题目：

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

题解：

双指针，i为s的index, j为p的index.

i 小于 s.length() 前提下检查 s.charAt(i) 和 p.charAt(j)是否match. 若是当前对应字符match, 双双后移一位。

若是不能match且p的当前字符是'*', 就更新星号的starIndex 和 最后检查的位置lastCheck. 

当再次遇到不match时，j回到startIndex的后一位，lastCheck后一一位，i到lastCheck位上.

出了while loop 说明s到头了，看j是否到头即可。但是若当前j是'*', 需要一致后移j.

Time Complexity: O(s.length() * p.length()).

Space: O(1).

AC Java:

public class Solution {
    public boolean isMatch(String s, String p) {
        if(p.length() == 0){
            return s.length() == 0;
        }
        int starIndex = -1; //标记星号的index
        int lastCheck = -1; //标记上次检查的位置
        
        int i = 0; 
        int j = 0;
        while(i<s.length()){
            if(j<p.length() && isMatch(s.charAt(i), p.charAt(j))){
                i++;
                j++;
            }else if(j<p.length() && p.charAt(j) == '*'){
                starIndex = j;
                lastCheck = i;
                j++;
            }else if(starIndex != -1){
                j = starIndex+1;
                lastCheck++;
                i=lastCheck;
            }else{
                return false;
            }
        }
        //s = "", p = "**"
        while(j<p.length() && p.charAt(j) == '*'){
            j++;
        }
        return j == p.length();
    }
    
    private boolean isMatch(char s, char p){
        return p == '?' || s == p;
    }
}

类似Regular Expression Matching.