Wildcard Matching

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

思想:用变量start记录*是否出现过；用ix和iy表示*可能比配的字符，即下一次需要匹配的开端

1. public boolean isMatch(String s, String p) {
2. int s1 = s.length();
3. int p1 = p.length();
4. if(s1==0) {
5. for(int i=0;i<p1;i++) {
6. if(p.charAt(i) != '*') {
7. return false;
8. }
9. }
10. return true;
11. }
12. boolean star = false;
13. int ix = 0;
14. int iy = 0;
15. int i= 0;
16. int j=0;
17. for(i=0,j=0;i<s1;i++,j++) { //完全匹配s1
18. char ptmp = '#'; //java中无结尾字符
19. if(j<p1)
20. ptmp = p.charAt(j);
21. switch(ptmp){
22. case '?':
23. break;
24. case '*':
25. ix = i;
26. iy = j;
27. star = true;
28. while(iy<p1 && p.charAt(iy) == '*') iy++;
29. if(iy==p1) return true;
30. i = ix - 1;
31. j = iy - 1;
32. break;
33. default:
34. if(ptmp!=s.charAt(i)) {
35. if(!star) {
36. return false;
37. }
38. ix++;
39. i = ix -1;
40. j = iy - 1;
41. }
42. }
43. }
44. while(j<p1 && p.charAt(j) == '*') j++;
45. if(j==p1) {
46. return true;
47. } else {
48. return false;
49. }
50. }

C代码：

1. bool isMatch(const char *s, const char *p) {
3. bool star = false;
4. const char *str,*ptr;
5. for(str = s,ptr=p;*str!='\0';str++,ptr++)
6. {
7. switch(*ptr)
8. {
9. case '?':
10. break;
11. case '*':
12. star=true;
13. s=str;
14. p=ptr;
15. while(*p=='*')
16. p++;
17. if(*p=='\0')
18. return true;
19. str=s-1;
20. ptr=p-1;
21. break;
22. default:
23. {
24. if(*str!=*ptr)
25. {
26. if(!star)
27. return false;
28. s++;
29. str=s-1;
30. ptr=p-1;
31. }
32. }
33. }
34. }
35. while(*ptr=='*')
36. ptr++;
37. return (*ptr=='\0');
38. }