第10题 正则表达式匹配（动态规划）

题目描述：

'.' 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

　　使用字符串p来表示字符串s，看是否匹配。比如“c*a*b“可以匹配”aab“，此时第一个*表示有0个c，第二个*表示有1个a。

 

思路：

1, If p.charAt(j) == s.charAt(i) :  dp[i][j] = dp[i-1][j-1];2, If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1];3, If p.charAt(j) == '*':    here are two sub conditions:               1   if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2]  //in this case, a* only counts as empty               2   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

比如：s="aaab" , p="c*a*b"

 c*a*ba10101a00000a00000b0000 

 

代码：

package T010;public class RegularExpressionMatching {    public static void main(String[] args) {        System.out.println(isMatch("aab","c*a*b"));    }    public static 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;            }        }        printArray(dp);        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]);                    }                }            }        }       // printArray(dp);        return dp[s.length()][p.length()];    }    /*      * 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     */    public static void printArray(boolean [] arr){        System.out.print("printArray:");        for(int i=0;i<arr.length;i++){            System.out.print(arr[i]+" ");        }        System.out.println("");    }        public static void printArray(boolean[][] V){        System.out.println("printArray:");        int rows = V.length;        int cols = V[0].length;        for(int i=0;i<rows;i++){            for(int j=0;j<cols;j++){                System.out.print(V[i][j]+" ");            }            System.out.println("");                    }        System.out.println("");    }}