WildCard Matching

lc 44

Implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character. '*' Matches any sequence of characters (including the empty sequence).

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", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false

dp解法 boolean[][] dp,

initial state; dp[0][0] = true; 当i为持续为＊的时候，dp[0][i + 1]= true;

状态转移： 如果是j 是？，dp[i + 1][j + 1] ＝ dp[i][j] 如果相同，dp[i + 1][j + 1] ＝ dp[i][j] 如果是“＊” ：

1. dp[i][j] = true || dp[i+1][j] = true, dp[i + 1][j + 1] = true;
2. 第三种交叉情况，但是要从k到i，只要找到一个是true，那么可以无限加string

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;
    int m = 0;
    if(p.length() == 1 && p.charAt(0) == '*') return true;
    while(m< p.length() - 1){
         if (p.charAt(m) == '*'  && p.charAt(m) == p.charAt(m + 1)){
               dp[0][m + 1] = true;
               dp[0][m + 2] = true;
             m++;

         }else {
             if(p.charAt(m) == '*'){
                   dp[0][m + 1] = true;
             }
             break;
         }
    }
    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) != '?') {
                    if(dp[i][j] || dp[i+1][j]) dp[i+1][j+1] = true;
                    else{
                        for(int k = 0; k <=i; k++){
                            if(dp[k][j]){
                                dp[i+1][j+1] = true;
                                break;
                            }
                        }
                    }
                    //dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1] || dp[i][j]);
              //  } else {
                 //   dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1] || dp[i][j]);
               // }
            }
        }
    }
    return dp[s.length()][p.length()];       
    }
}

lc10 Regular Expression Matching

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)

意思是match 0 或者多个先前的word

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

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]) ){
            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] || dp[i][j]);
                }
            }
        }
    }
    return dp[s.length()][p.length()];
}