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") → 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

　　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	*	b
a	1	0	1	0	1
a	0	0	0	0	0
a	0	0	0	0	0
b	0	0	0	0

 

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("");
    }
}