LeetCode Regular Expression Matching

Description:

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).The function prototype should be:bool isMatch(const char *s, const char *p)Some examples:isMatch("aa","a") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

Solution:

dp[i][j] represents whether s[0,i-1] and p[0,j-1] is a match.

We can have three situations s[i-1] and p[j-1]:

if p[j] != '.' and [j] != '*'    dp[i][j] = dp[i-1][j-1] + s[i-1] == p[j-1];else if p[j] == '.'    dp[i][j] = dp[i-1][j-1];else // p[j] = '*'    if (j>1)        if (dp[i][j-1] || dp[i][j-2])            dp[i][j] = true        else if(i>0 && dp[i-1][j] && (p[j-2]=='.' || s[i-1]==p[j-2] ) )  //(1)            dp[i][j] = true

(1) Because we may have the situation that "" and ".*" is a match, so we need to start i from 0.

<span style="font-size:18px;">public class Solution {public boolean isMatch(String s, String p) {if (s == null || p == null)return false;int n = s.length();int m = p.length();boolean dp[][] = new boolean[n + 1][m + 1];dp[0][0] = true;char ch1, ch2;for (int i = 0; i <= n; i++) {for (int j = 1; j <= m; j++) {ch2 = p.charAt(j - 1);if (ch2 == '*') {if (j > 1) {if (dp[i][j - 1] || dp[i][j - 2])dp[i][j] = true;else if (i > 0&& (p.charAt(j - 2) == s.charAt(i - 1) || p.charAt(j - 2) == '.') && dp[i - 1][j])dp[i][j] = true;}} else if (i > 0) {ch1 = s.charAt(i - 1);if (ch2 == '.') {dp[i][j] = dp[i - 1][j - 1];} else {if (ch1 == ch2)dp[i][j] = dp[i - 1][j - 1];}}}}return dp[n][m];}}</span>