java面试题（4）

原题：

// 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

答案:

public class Solution {    public boolean isMatch(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) == '.') {                    dp[i + 1][j + 1] = dp[i][j];                }                if(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()];    }}