leetcode Wildcard Matching greedy algrithm

The recursive program will result in TLE like this:

 

class Solution {
 public:
  bool isMatch(const char *s, const char *p) {
    // IMPORTANT: Please reset any member data you declared, as
    // the same Solution instance will be reused for each test case.
    if (*s == *p && *s == '\0')
      return true;
    if (*p == '?' || *s == *p)
      return isMatch(s + 1, p + 1);
    else if (*p == '*') {
      int i;
      for (i = 0; *(s + i); ++i)
        if (isMatch(s + i, p + 1))
          return true;
      if (isMatch(s + i, p + 1))
        return true;
      
      return false;
    }
    else if (*p != *s)
      return false;
  }
};

So it's necessary to write an non-recursive program. The key point is to match the '*' in p string. We could attempt to match '*' with 0...n characters in s, i.e., the character after '*' maybe match any position in s regardless a series of characters in s. Take notice that consecutive '*'s are equal to one '*'. Based on that, a lengthy code is written as :

 

 

class Solution {
 public:
  bool isMatch(const char *s, const char *p) {
    // IMPORTANT: Please reset any member data you declared, as
    // the same Solution instance will be reused for each test case.
    bool star = false, staremerge = false;
    const char *str = s, *ptr = p, *ss = s, *pp = p;
    for (str = ss, ptr = pp; *str && *ptr || *str == '\0' && *ptr == '*'; ++str, ++ptr) {
      if (*ptr == '*') {
        star = staremerge = true;
        while (*ptr == '*')
          ++ptr;
        if (*ptr == '\0')
          return true;
        ss = str;
        pp = ptr;
        --str;
        --ptr;
      }  
      else {
        if (!star) {
          if( !staremerge ) {
            if (*str != *ptr && *ptr != '?' ||*(ptr + 1) == '\0' && *(str + 1) != '\0')  
              return false;
          }
          else {
            if (*str != *ptr && *ptr != '?' ||*(ptr + 1) == '\0' && *(str + 1) != '\0') {
              str = ss++;
              ptr = pp - 1;
              star = true;
            }
              
          }
        }
        else if (star) {
          if ( *str != *ptr && *ptr != '?') {
            ss = str + 1;  
            --ptr;
          }
          else {
            star = false;
            if (*(ptr + 1) == '\0' && *(str + 1) != '\0') {
              str = ss++;
              ptr = pp - 1;
              star = true;
            }            
          }
        }
      }
    }
    if (*str == *ptr && *str == '\0')
      return true;
    else
      return false;
  }
};

Some suggestions about this code: 

 

1. There is no need to refresh the status of star, staremerge. Only one star is enough, because the character(for example, 'a') always needs to match some 'a' in s. Matching the former 'a' is better than the latter 'a' in s as is illustrated in the figure. I.e., there is no need to record the matching range for every '*', the latest '*' has the largest range of choice. 

2. sbegin is refreshed when mismatch occurs and pbegin is refreshed when meeting new '*';

3. Focusing on s is better than handling the two strings at the same time. 

So the final concise code is like:

 

class Solution {
 public:
  bool isMatch(const char *s, const char *p) {
    // IMPORTANT: Please reset any member data you declared, as
    // the same Solution instance will be reused for each test case.
    const char *sbegin = s, *pbegin = p, *str = s, *ptr = p;
    bool star = false;
    for (str = s, ptr = p; *str || *ptr == '*'; ++str, ++ptr) {
      if (*ptr == '*') {
        star = true;
        while (*ptr == '*') 
          ++ptr;
        if (*ptr == '\0')
          return true;
        pbegin = ptr--;
        sbegin = str--;
      }
      else if (*str != *ptr && *ptr != '?'){
        if (!star)
          return false;
        str = sbegin++;
        ptr = pbegin - 1;
      }
    }
    return *ptr == '\0';  
  }
};

 

 

转载于:https://www.cnblogs.com/james1207/p/3395436.html