10.Regular-Expression-Matching

题目地址

https://leetcode.com/problems/regular-expression-matching/

题目描述

``````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 preceding 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``````

代码

Approach 1: Recursion

``````public boolean isMatch(String text, String pattern) {
if (pattern.isEmpty()) return text.isEmpty();

boolean first_match = (!text.isEmpty()
&& (pattern.charAt(0) == text.charAt(0)
|| pattern.charAt(0) == '.'));
//只有长度大于 2 的时候，才考虑 *
if (pattern.length() >= 2 && pattern.charAt(1) == '*') {
//两种情况
//pattern 直接跳过两个字符。表示 * 前边的字符出现 0 次
//pattern 不变，例如 text = aa ，pattern = a*，第一个 a 匹配，然后 text 的第二个 a 接着和 pattern 的第一个 a 进行匹配。表示 * 用前一个字符替代。
return (isMatch(text, pattern.substring(2)) ||
(first_match && isMatch(text.substring(1), pattern)));
} else {
return first_match && isMatch(text.substring(1), pattern.substring(1));
}
}``````

Approach #2 DP - Bottom - up

dp[ i ] [ j ] = 表示 text 从 i 开始到最后，pattern 从 j 开始到最后，此时 text 和 pattern 是否匹配。

``````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 + 1) == '*':
here are two sub conditions:
dp[i][j] = first_match && dp[i + 1][j]       // in this case, a* counts as multiple a
dp[i][j] = first_match && dp[i + 1][j + 1]; // in this case, a* counts as single a
dp[i][j] = dp[i][j + 2]    // in this case, a* counts as empty``````
``````public boolean isMatch(String text, String pattern) {
// 多一维的空间，因为求 dp[len - 1][j] 的时候需要知道 dp[len][j] 的情况，
// 多一维的话，就可以把 对 dp[len - 1][j] 也写进循环了
boolean[][] dp = new boolean[text.length() + 1][pattern.length() + 1];
// dp[len][len] 代表两个空串是否匹配了，"" 和 "" ，当然是 true 了。
dp[text.length()][pattern.length()] = true;

// 从 len 开始减少
for (int i = text.length(); i >= 0; i--) {
for (int j = pattern.length(); j >= 0; j--) {
// dp[text.length()][pattern.length()] 已经进行了初始化
if (i == text.length() && j == pattern.length()) continue;

boolean first_match = (i < text.length() && j < pattern.length()
&& (pattern.charAt(j) == text.charAt(i)
|| pattern.charAt(j) == '.'));
if (j + 1 < pattern.length() && pattern.charAt(j + 1) == '*') {
// 跳过pattern || 匹配一个，继续往下匹配该pattern
dp[i][j] = dp[i][j + 2] || first_match && dp[i + 1][j];
} else {
// 逐个匹配
dp[i][j] = first_match && dp[i + 1][j + 1];
}
}
}
return dp[0][0];
}``````

Approach #3 Dynamic Programming

`dp[i][j]` denotes if s.substring(0,i) is valid for pattern `p.substring(0,j)`.

For example `dp[0][0] == true` (denoted by y in the matrix) because when s and p are both empty they match. So if we somehow base `dp[i+1][j+1]` on previos `dp[i][j]`'s then the result will be `dp[s.length()][p.length()]`

``````class Solution {
public boolean isMatch(String s, String p) {
if (p == null || p.length() == 0) return (s == null || s.length() == 0);

boolean dp[][] = new boolean[s.length() + 1][p.length() + 1];
dp[0][0] = true;
for (int j = 1; j < p.length(); j += 2) {
// 匹配空串，a*b*c*.. 各匹配0次
dp[0][j + 1] = p.charAt(j) == '*' && dp[0][j - 1];
}
// 从 (1, 1) 开始
for (int j = 1; j <= p.length(); j++) {
for (int i = 1; i <= s.length(); i++) {
if (p.charAt(j - 1) == s.charAt(i - 1) || p.charAt(j - 1) == '.') {
// 各移动比较一次
dp[i][j] = dp[i - 1][j - 1];
} else if(p.charAt(j - 1) == '*') {
// 0 次匹配 || 匹配一次（first match && 前i-1匹配）
dp[i][j] = dp[i][j - 2] || ((s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') && dp[i - 1][j]);
}
}
}

return dp[s.length()][p.length()];
}
}``````

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