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
分析：
动态规划，dp[i][j]表示s[0]到s[i]和p[0]到p[j]是否match。这里动态转移方程比较复杂
if (p[j]==’.’||p[j]==s[i]) dp[i+1][j+1]=dp[i][j]
if(p[j]==’*’)
if(p[j-1]!=s[i]) dp[i+1][j+1]=dp[i][j-1] //in this case, a* only counts as empty
if p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == ‘.’:
dp[i][j] = dp[i-1][j] //in this case, a* counts as multiple a
or dp[i][j] = dp[i][j-1] // in this case, a* counts as single a
or dp[i][j] = dp[i][j-2] // in this case, a* counts as empty
class Solution {
public:
bool isMatch(string s, string p) {
vector<vector<int>> dp(s.size()+1, vector<int> (p.size()+1, 0));
dp[0][0]=1;
for(int i=0; i<p.size(); i++)
if(p[i]=='*'&&dp[0][i-1]==1)
dp[0][i+1]=1;
for(int i=0; i<s.size(); i++)
for(int j=0; j<p.size(); j++){
if(p[j]=='.')
dp[i+1][j+1]=dp[i][j];
else if(s[i]==p[j])
dp[i+1][j+1]=dp[i][j];
else if(p[j]=='*'){
if(p[j-1]!= s[i] && p[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]);
}
}
if(dp[s.size()][p.size()]==1)
return true;
else
return false;
}
};