[LintCode] 通配符查询

动态规划：

 1 class Solution {
 2 public:
 3     /**
 4      * @param s: A string 
 5      * @param p: A string includes "?" and "*"
 6      * @return: A boolean
 7      */
 8     bool isMatch(const char *s, const char *p) {
 9         // write your code here
10         int m = strlen(s), n = strlen(p);
11         vector<bool> cur(m + 1, false);
12         cur[0] = true;
13         for (int j = 1; j <= n; j++) {
14             bool pre = cur[0];
15             cur[0] = cur[0] && p[j - 1] == '*';
16             for (int i = 1; i <= m; i++) {
17                 bool temp = cur[i];
18                 if (p[j - 1] != '*')
19                     cur[i] = pre && (s[i - 1] == p[j - 1] || p[j - 1] == '?');
20                 else cur[i] = cur[i - 1] || cur[i];
21                 pre = temp;
22             }
23         }
24         return cur[m];
25     }
26 };

贪心：

 1 class Solution {
 2 public:
 3     /**
 4      * @param s: A string 
 5      * @param p: A string includes "?" and "*"
 6      * @return: A boolean
 7      */
 8     bool isMatch(const char *s, const char *p) {
 9         // write your code here
10         const char* asterisk = NULL;
11         const char* match = NULL;
12         while (*s) {
13             if (*p == '*') {
14                 asterisk = p++;
15                 match = s;
16             }
17             else if (*s == *p || *p == '?') {
18                 s++;
19                 p++;
20             }
21             else if (asterisk) {
22                 s = ++match;
23                 p = asterisk + 1;
24             }
25             else return false;
26         }
27         while (*p == '*') p++;
28         return !*p;
29     }
30 };

 

posted @ 2015-07-06 12:06  jianchao-li  阅读(219)  评论(0)  编辑  收藏  举报