L
L
LintCode & LeetCode
Search
⌃K
Comment on page

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:
  1. 1.
    if p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1]
  2. 2.
    if p.charAt(j) == '.': dp[i][j] = dp[i - 1][j-1]
  3. 3.
    if p.charAt(j) == '*' : sub conditions
    1. 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. 2.
      if p.charAt(j-1) == s.charAt(i) or p.charAt(j - 1) == '.':
      1. 1.
        dp[i][j]= dp[i-1][j] // in this case, a* counts as multiple a
      2. 2.
        (REDUNDANT) or dp[i][j] = dp[i][j-1] // in this case, a* counts as single a
      3. 3.
        or dp[i][j] = dp[i][j-2] // in this case, a* counts as empty
Answer:
dp[m][n]
另：中文解释 From @caiconghuai GitHub:
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

Previous
Maximal Square
Next
Wildcard Matching
Last modified 3yr ago