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# 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
`dp[m][n]`

• 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()];
}
}