Regular Expression Matching

Given an input string (s) and a pattern (p), 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).

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 = "a*"

Output:
 true

Explanation:
 '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:

s = "ab"
p = ".*"

Output:
 true

Explanation:
 ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:

s = "aab"
p = "c*a*b"

Output:
 true

Explanation:
 c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:

s = "mississippi"
p = "mis*is*p*."

Output:
 false

Analysis

Two string / array comparison, and it has subproblems that could lead to Dynamic Programming solution.

isMatch(char[] str, char[] pattern)

State:

dp[i][j]

Defined as new boolean[str.length + 1][pattern.length + 1]

The first i characters in the str[] matches first j characters in pattern[]

Note: the corresponding index (subscript) will be i - 1 in str[], j - 1 in pattern[]

Initialization:

Assumption: '*' will never be the first character

dp[0][0] = true - no chars chosen from either of the string / char array, should match

then for dp[0][i] (for a* case):

for (int i = 1; i <= pattern.length; i++) {
    if (pattern[i-1] == '*') {
        dp[0][i] = dp[0][i-2];
    }    
}

State Transfer Function:

if p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1]
if p.charAt(j) == '.': dp[i][j] = dp[i - 1][j-1]
if p.charAt(j) == '*' : 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(j-1) == s.charAt(i) or p.charAt(j - 1) == '.':
  1. dp[i][j]= dp[i-1][j] // in this case, a* counts as multiple a
  2. (REDUNDANT) or dp[i][j] = dp[i][j-1] // in this case, a* counts as single a
  3. or dp[i][j] = dp[i][j-2] // in this case, a* counts as empty

Answer:

dp[m][n]

在正则表达式中，由于不同的字符会有不同的匹配规则，其中.和*比较特别，需要对这两类字符进行分类讨论。
定义状态
dp[i][j] - 表示输入串长度为i，模式串长度为j时，是否能匹配。
初始化状态值：
输入串为空，模式串为空：dp[0][0] 必然可以匹配，即 dp[0][0]=true
输入串非空，模式串为空：由于模式串为空，此时必然不可匹配，即dp[i][0]=false, i > 0
输入串为空，模式串非空：此时如果模式串中有*，则是可以匹配0个前面的字符的，所有我们通过条件 p.charAt(j-1) == '*' && dp[0][j-2] 来判断是否可匹配
转移方程，转移方程表示通过已知的状态值来求解未知的状态值，在这里就是我们要求解
dp[i][j]，但是对于所有长度小于i的输入串和长度小于j的模式串，我们已知其匹配情况。现在，我们怎么利用前面的信息，来得到dp[i][j]的值。我们首先需要分析模式串的当前字符来做判断:
p.charAt(j-1) == s.charAt(i-1)当前模式串字符一样，可以将当前字符匹配掉，然后状态转化如下: dp[i][j] = dp[i-1][j-1];
p.charAt(j-1) == '.'当前模式串字符为.，可以匹配任意的字符，可以将当前输入串字符匹配掉，然后状态转化为:dp[i][j] = dp[i-1][j-1]
p.charAt(j-1) == '*'由于*可以匹配0个或多个它的上一个字符，所以根据上一个字符的不同还需要做不同的分析：
p.charAt(j-2) != s.charAt(i-1) && p.charAt(j-2) != '.'，这种情况说明上一个字符即不是.，也不和输入串的当前字符一样，所有无法用一个或多个上一个字符对输入串进行匹配，此时*只能匹配0个上一个字符: dp[i][j] = dp[i][j-2];
如果是这个条件里，说明上一个字符和输入串的当前字符是可以进行匹配的，故此时*即可以匹配0个前一个字符，也可以匹配1个前一个字符，也可以匹配多个前一个字符: dp[i][j] = (dp[i][j-1] || dp[i-1][j] || dp[i][j-2]) Note: 这种情况，也可以用dp[i][j] = (dp[i-1][j] || dp[i][j-2])

Solution

DP Solution (12ms 97% AC)

class Solution {
    public boolean isMatch(String s, String p) {
        char[] text = s.toCharArray();
        char[] pattern = p.toCharArray();
        int m = text.length;
        int n = pattern.length;
        boolean dp[][] = new boolean[m + 1][n + 1];

        dp[0][0] = true;
        // Deals with patterns like a* or a*b* or a*b*c*
        for (int i = 1; i < n + 1; i++) {
            // Assumption: '*' will never be the first character
            if (pattern[i-1] == '*') {
                dp[0][i] = dp[0][i - 2];
            }
        }

        for (int i = 1; i < m + 1; i++) {
            for (int j = 1; j < n + 1; j++) {
                if (pattern[j - 1] == '.' || pattern[j - 1] == text[i - 1]) {
                    dp[i][j] = dp[i-1][j-1];
                } else if (pattern[j - 1] == '*')  {
                    dp[i][j] = dp[i][j - 2];
                    if (pattern[j-2] == '.' || pattern[j - 2] == text[i - 1]) {
                        dp[i][j] = dp[i][j] | dp[i - 1][j];
                    }
                } else {
                    dp[i][j] = false;
                }
            }
        }
        return dp[m][n];
    }
}

DP - Reference

class Solution {
    //dp[i][j] first ith characters match jth characters
    // abc
    // ab.
    // ab*
    //initial dp[0][0] = true; dp[][0] = false;
    public boolean isMatch(String s, String p) {
        if(s == null || p == null){
            return false;
        }
        boolean[][] dp = new boolean[s.length() +1 ][p.length() + 1];
        for(int i = 0; i <= s.length(); i++){
            for(int j = 0;j <= p.length(); j++){
                if(i == 0 && j == 0){
                    dp[i][j] = true;
                    continue; //这个地方不能少了continue
                }
                if(j == 0){
                    dp[i][j] = false;
                    continue;//contiue不能少了
                }
                //dp[i][j] = false; // 默认false 不能没了
                if(p.charAt(j - 1) != '*'){
                    if(i > 0 && (p.charAt(j-1) == '.' || p.charAt(j-1) == s.charAt(i-1)))
                        dp[i][j] |= dp[i-1][j-1];
                }
                else{
                    if(j > 1)
                        dp[i][j] |= dp[i][j-2];
                    if(i > 0 && (p.charAt(j-2) == '.' || s.charAt(i-1) == p.charAt(j-2))){
                        dp[i][j] |= dp[i-1][j];
                    }
                }               
            }
        }
        return dp[s.length()][p.length()];
    }
}

Reference

PreviousMaximal Square NextWildcard Matching

Last updated 5 years ago

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