KMP½Ž—法
KMP½Ž—法是一¿U改˜q›çš„å—符串匹é…算法，æ¤ç®—法å¯ä»¥åœ¨O(n+m)的时间数é‡çñ”上完æˆä¸²çš„模å¼åŒ¹é…æ“作，基本æ€æƒ³æ˜¯ï¼Œæ¯å½“匚w…˜q‡ç¨‹ä¸å‡ºçŽ°å—½W¦ä¸²æ¯”较ä¸ç‰æ—Óž¼Œä¸éœ€å›žæº¯æŒ‡é’ˆåQŒè€Œæ˜¯åˆ©ç”¨å·²ç»å¾—到çš?#8220;部分匚w…”¾l“æžœž®†æ¨¡å¼å‘å?#8220;滑动”ž®½å¯èƒ½è¿œçš„一ŒDµè·¼›»ï¼Œæ®å¾˜q›è¡Œæ¯”较ã€?/font>
定义åQ?/font>
①　è¦æœç´¢çš„关键å—符ä¸?nbsp;key K1K2......Kmã€?/font>
②　被æœç´¢çš„æºå—½W¦ä¸² source S1S2.....Snã€?/font>
u 针对è¦æŸ¥æ‰„¡š„关键å—å—½W¦ä¸²æž„é€ å¤±æ•ˆå‡½æ•°f(s)åQŒè¯¥å‡½æ•°çš„功能就是在比较å½?/font>Ki != Sj的时候计½Ž—出从K的第多少个嗽W¦å¼€å§‹é‡æ–°æ¯”较ã€?/font>
使得K1K2...Kf(s)是最长的既是K1K2...Ks的的真剾~€åQŒåˆæ˜?/font>K1K2...Ks的厾~€çš„å—ä¸?/font>。也ž®±æ˜¯è¯´å¦‚果我们试囄¡”¨ä¸€ä¸ªæ–‡æœ¬ä¸²x匚w…K1K2......KmåQ?/font>òq¶ä¸”我们已ç»åŒšw…了å‰i个嗽W¦ï¼Œä½†åŒ¹é…Ki+1的时候失败，也就是说x的下一个佾|®ä¸æ˜¯Ki+1åQŒé‚£ä¹ˆf(s)ž®±æ˜¯å¯èƒ½å’Œæˆ‘们当å‰ä½¾|®äØ“¾l“尾的文本串匚w…的最长的K1K2......Km的剾~€å’ŒåŽ¾~€y。也ž®±æ˜¯è¯´æ–‡æœ?/font>x的下一个佾|®å’ŒKf(s)比较åQ?/font>X当å‰ä½ç½®ä¹‹å‰çš„f(s)个嗽W¦å’ŒK1..Kf(s)是匹é…的。所以直接从Kf(s)之厘q›è¡Œæ¯”较ã€?/font>
½Ž—法如下åQ?/font>
|
t = 0; f(1) = 0; for (i = 1; i < m; i++) { while(t > 0 && (Ki+1 != Kt+1)) t = f(t); if(Ki+1 == Kt+1) { t = t + 1; f(i+1) = t; } else f(i+1) = 0; } |
对于ä¸?/font>"a b a b a a"的计½Ž—f(s)的过½E‹å¦‚下：
|
s |
f(s) |
当å‰å—符ä¸?/span> |
说明 |
|
1 |
0 |
a |
æ— çœŸå‰ç¼€ |
|
2 |
0 |
ab |
æ— çœŸå‰ç¼€ |
|
3 |
1 |
aba |
"a"æ˜?aba"的真å‰ç¼€å’ŒåŽ¾~€ |
|
4 |
2 |
abab |
"ab"æ˜?abab"的真å‰ç¼€å’ŒåŽ¾~€ |
|
5 |
3 |
ababa |
"aba"æ˜?ababa"的真å‰ç¼€å’ŒåŽ¾~€ |
|
6 |
1 |
ababaa |
"a"æ˜?ababaa"的真å‰ç¼€å’ŒåŽ¾~€ |
u 针检æŸ?/font>S1S2.....Sn是å¦åŒ…å«å…³é”®å?/font>K1K2......Kmçš„ç®—æ³?/font>ã€?/font>
|
s = 0; for (i = 1; i <= n i++) { // ä¸‹æ ‡ä»?å¼€å§?/font> while (s > 0 && Si != Ks+1) s = f(s); if(Si == Ks+1) s = s + 1; if(s == n) return "yes"; } return "no"; |
helloworld2008 2010-03-21 01:14 å‘表评论
]]>2 public class ComputerWeek {
3
4 /**
5 * 计算æŸä¸€ä¸ªæ—¥æœŸæ˜¯æ˜ŸæœŸå‡?br /> 6 * @param year òq?br /> 7 * @param month æœ?br /> 8 * @param day æ—?br /> 9 * @return 星期
10 */
11 public int getWeek(int year, int month, int day) {
12 // 首先我们知é“2008-01-07是星期一
13 int weeks[] = {1,2,3,4,5,6,7};
14 int standardDay = getDaysFromAD(2008, 1, 7);
15 int temps = getDaysFromAD(year,month,day);
16 temps = temps - standardDay;
17 if (temps >= 0) {
18 temps %= 7;
19 return weeks[temps];
20 } else {
21 temps %=7;
22 temps += 7;
23 temps %= 7;
24 return weeks[temps];
25 }
26 }
27
28 /**
29 * 计算从公å…?òq?æœ?å?nbsp;到这个日期的天数
30 * @param year
31 * @param month
32 * @param day
33 * @return
34 */
35 public int getDaysFromAD(int year, int month, int day) {
36 int days = 0;
37 for (int i = 1; i < year; i++) {
38 if (isLeapYear(i)) {
39 days += 366;
40 } else {
41 days += 365;
42 }
43 }
44 for (int i = 1; i < month; i++) {
45 if (i == 1 || i == 3 || i == 5 || i == 7 || i == 8 || i == 10 || i == 12) {
46 days += 31;
47 } else if (i == 2) {
48 if (isLeapYear(year)) {
49 days += 29;
50 } else {
51 days += 28;
52 }
53 } else {
54 days += 30;
55 }
56 }
57 days += day;
58 return days;
59 }
60
61 /**
62 * 判æ–是å¦ä¸ºé—°òq?br /> 63 * @param year
64 * @return
65 */
66 public boolean isLeapYear(int year) {
67 if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)) {
68 return true;
69 } else {
70 return false;
71 }
72 }
73
74 public static void main(String args[]) {
75 System.out.println(new ComputerWeek().getWeek(2008,12,31));
76 }
77 }
78
helloworld2008 2008-01-10 16:54 å‘表评论
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