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") → false isMatch("aa","aa") → true isMatch("aaa","aa") → false isMatch("aa", "a*") → true isMatch("aa", ".*") → true isMatch("ab", ".*") → true isMatch("aab", "c*a*b") → true
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正则表达式判断，这个题一眼看上去就是dp问题，不过想要跑通还是挺复杂的，卡了很久才通过，在这里贴出代码以便复习：
public class Solution { public boolean isMatch(String s, String p) { //动态规划版本 if(s==null||p==null){ return false; } boolean[][]res = new boolean[s.length()+1][p.length()+1]; //动态规划表 res[0][0] = true; for(int i=0;i<p.length();i++){ //初始化第一行 s为空时 p各个长度的判断值 if(p.charAt(i)=='*'&&res[0][i-1]){ res[0][i+1] = true; } } for(int i=0;i<s.length();i++){ for(int j=0;j<p.length();j++){ if(s.charAt(i)==p.charAt(j)||p.charAt(j)=='.'){ //当s[i]==p[j]或者p[j]=='.'直接看i-1,j-1的位置 res[i+1][j+1] = res[i][j]; } if(p.charAt(j)=='*'){ if(s.charAt(i)!=p.charAt(j-1)&&p.charAt(j-1)!='.'){ //当p[j]=='*'，如果前一个值和s[i]不等，则只有*取0的判断可能成立 res[i+1][j+1] = res[i+1][j-1]; } else{ res[i+1][j+1] = res[i+1][j-1]||res[i+1][j]||res[i][j+1]; //p[j-1]==s[i]这时*可能是0，1，多个 } } } } return res[s.length()][p.length()]; } }