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 = "cab"
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 = "misisp*."
Output: false
该题是正则表达式匹配，要求
'.' 匹配任意单个字符 '*' 匹配零个或多个前面的元素。
判断两个字符串是否正则表达式匹配。
如果没有 * 问题会更容易，我们只需从左到右检查文本的每个字符是否与模式匹配。
当存在 * 时，我们可能需要检查文本的许多不同后缀，看它们是否与模式的其余部分匹配。 递归解决方案是表示这种关系的直接方式。不断调用str.substring方法获取子串，并递归调用方法。
递归方法
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) == '.'));
if (pattern.length() >= 2 && pattern.charAt(1) == '*') {
return (isMatch(text, pattern.substring(2)) || (first_match && isMatch(text.substring(1), pattern)));
} else {
return first_match && isMatch(text.substring(1), pattern.substring(1));
}
}
动态规划
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()];
}