string T = "Rabin–Karp string search algorithm: Rabin-Karp"; string P = "Rabin"; int q = 101; // A prime number int d = 16; Rabin_Karp_search(T, P,d,q); system("pause"); return 0; } 參考资料: 《算法导论》 http://www.geeksforgeeks.org/searching-for-patterns-set-3-rabin...
intmain() { string p,s; p="Rabin"; s="Rabin–Karp string search algorithm: Rabin-Karp"; int m=101;//素数 int base=26;//基数,这里取26好了 Rabin_Karp(p,s,base,m); return0; } 自身匹配问题 给定一个长度为n的串s,求其子串中是否存在相同的且长度都为l的串,若存在,输出其出现次数以...
// 在文本串 txt 中搜索模式串 pat 的起始索引int search(String txt, String pat) {int N = txt.length(), L = pat.length();for (int i = 0; i + L <= N; i++) {String subStr = txt.substring(i, i + L);if (subStr.equals(pat)){// 在 txt 中找到模式串 pat,返回起始索引retur...
Rabin–Karp algorithmJump to: navigation, search The Rabin–Karp algorithm is a randomized algorithm for the string search problem that finds all probable matches for the needle in the haystack in linear time. Together with the use of a hash table, it can also find all probable matches for ...
Previous String: ins First Letter of Previous String: i(9) New Letter: a(1) New String: "nsa" 2462 - 9 = 2453 2453 / 11 = 223 223 + 1 X 11² = 344 这是一个匹配! 现在我们将模式与当前字符串进行比较。由于两个字符串都匹配,因此子字符串存在于此字符串中。然后我们返回子字符串的起...
String subStr = s.substring(i, i +10);if(seen.contains(subStr)){//之前出现过,找到一个重复的res.add(subStr); }else{//之前没出现过,加入集合seen.add(subStr); } }returnnewLinkedList<>(res); } 这个算法肯定是没问题的,只是时间复杂度略高。假设s的长度为N,目标子串的长度为L(本题L = 10)...
// primeRK is the prime base used in Rabin-Karp algorithm.constprimeRK=16777619// hashStr returns the hash and the appropriate multiplicative// factor for use in Rabin-Karp algorithm.funchashStr(sep string)(uint32,uint32){hash:=uint32(0)fori:=0;i<len(sep);i++{hash=hash*primeRK+uint...
// factor for use in Rabin-Karp algorithm. func hashStr(sep string) (uint32, uint32) { hash := uint32(0) charcode := [...]uint32{5,2,0} for i := 0; i < len(sep); i++ { //hash = hash*primeRK + uint32(sep[i]) ...
import java.math.BigInteger; import java.util.Random; /** * The {@code RabinKarp} class finds the first occurrence of a pattern string * in a text string. * * This implementation uses the Rabin-Karp algorithm. * * For additional documentation, * see Section 5.3 of * Algorithms...
Previous String: ins First Letter of Previous String: i(9) New Letter: a(1) New String: "nsa" 2462 - 9 = 2453 2453 / 11 = 223 223 + 1 X 11² = 344 这是一个匹配! 现在我们将模式与当前字符串进行比较。由于两个字符串都匹配,因此子字符串存在于此字符串中。然后我们返回子字符串的...