Implement regular expression matching with support for '.' and '*'.

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

思想：

可以想象成走二维格子的问题，从左上角开始找，找符合要求的格子，上一个格子影响下一个格子

1, If p.charAt(j) == s.charAt(i) :  dp[i][j] = dp[i-1][j-1];//符合要求则走斜线2, If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1];//符合要求走斜线3, If p.charAt(j) == '*':    存在两种可能:               1   if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2]  //这种情况下，不符合条件               2   if p.charAt(j-1) == s.charAt(i) or p.charAt(j-1) == '.':                              dp[i][j] = dp[i-1][j]    //这种情况下存在多个                            or dp[i][j] = dp[i][j-1]   // 这种情况下存在一个                           or dp[i][j] = dp[i][j-2]   // 这种情况下不符合条件

代码如下：

public class RegularExpressionMatching {public boolean isMatching(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) == '.' || 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()];}}