leetcode 10. Regular Expression Matching

题目

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


解析
当模式中的第二个字符不是“*”时：
1、如果字符串第一个字符和模式中的第一个字符相匹配，那么字符串和模式都后移一个字符，然后匹配剩余的。
2、如果 字符串第一个字符和模式中的第一个字符相不匹配，直接返回false。


而当模式中的第二个字符是“*”时：
如果字符串第一个字符跟模式第一个字符不匹配，则模式后移2个字符，继续匹配。如果字符串第一个字符跟模式第一个字符匹配，可以有3种匹配方式：
1、模式后移2字符，相当于x*被忽略；
2、字符串后移1字符，模式后移2字符；
3、字符串后移1字符，模式不变，即继续匹配字符下一位，因为*可以匹配多位；

public class Solution {
    public boolean match(char[] str, char[] pattern)
    {
        if (str == null || pattern == null)
            return false;
        int strIndex = 0;
        int patternIndex = 0;
        return matchCore(str, strIndex, pattern, patternIndex);
    }
 
    public boolean matchCore(char[] str, int strIndex, char[] pattern, int patternIndex) {
        if (str.length == strIndex && patternIndex == pattern.length)
            return true;
        if (strIndex != str.length && pattern.length == patternIndex)
            return false;
 
        //模式第二个是*，且字符串第一个和模式第一个匹配，分三种情况；否则模式后移。
        if (patternIndex + 1 < pattern.length && pattern[patternIndex + 1] == '*') {
            if (strIndex != str.length && str[strIndex] == pattern[patternIndex]
               || (pattern[patternIndex] == '.' && strIndex != str.length)) {
                return matchCore(str, strIndex, pattern, patternIndex + 2)
                        || matchCore(str, strIndex + 1, pattern, patternIndex + 2)
                        || matchCore(str, strIndex + 1, pattern, patternIndex);
            } else
                return matchCore(str, strIndex, pattern, patternIndex + 2);
        }
 
 
        //模式第2个不是*，且字符串第1个跟模式第1个匹配，则都后移1位，否则直接返回false
        if ((strIndex != str.length && str[strIndex] == pattern[patternIndex])
                || (strIndex != str.length && pattern[patternIndex] == '.')) {
            return matchCore(str, strIndex + 1, pattern, patternIndex + 1);
        }
        return false;
    }
}

This Solution use 2D DP. beat 90% solutions, very simple.

Here are some conditions to figure out, then the logic can be very straightforward.

1, If p.charAt(j) == s.charAt(i) :  dp[i][j] = dp[i-1][j-1];
2, If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1];
3, If p.charAt(j) == '*': 
   here are two sub conditions:
               1   if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2]  //in this case, a* only counts as empty
               2   if p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == '.':
                              dp[i][j] = dp[i-1][j]    //in this case, a* counts as multiple a 
                           or dp[i][j] = dp[i][j-1]   // in this case, a* counts as single a
                           or dp[i][j] = dp[i][j-2]   // in this case, a* counts as empty

public boolean isMatch(String s, String p) {

    if (s == null || p == null) {
        return false;
    }
    boolean[][] dp = new boolean[s.length()+1][p.length()+1];
    dp[0][0] = true;
    for (int i = 0; i < p.length(); i++) {
        if (p.charAt(i) == '*' && dp[0][i-1]) {
            dp[0][i+1] = true;
        }
    }
    for (int i = 0 ; i < s.length(); i++) {
        for (int j = 0; j < p.length(); j++) {
            if (p.charAt(j) == '.') {
                dp[i+1][j+1] = dp[i][j];
            }
            if (p.charAt(j) == s.charAt(i)) {
                dp[i+1][j+1] = dp[i][j];
            }
            if (p.charAt(j) == '*') {
                if (p.charAt(j-1) != s.charAt(i) && p.charAt(j-1) != '.') {
                    dp[i+1][j+1] = dp[i+1][j-1];
                } else {
                    dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1]);
                }
            }
        }
    }
    return dp[s.length()][p.length()];
}