KMP算法实现Python/Java
2018-03-31 本文已影响0人
蛮三刀酱
kmp算法的核心时间复杂度就是O(m+n)
参考
原理:
http://www.ruanyifeng.com/blog/2013/05/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm.html
Java:http://blog.csdn.net/christ1750/article/details/51259425
Python:http://blog.csdn.net/handsomekang/article/details/40978213
Python
#KMP
def kmp_match(self, s, p):
m = len(s)
n = len(p)
cur = 0 # 起始指针cur
table = self.partial_table(p)
# print(table)
while cur <= m-n: # 长度不够就终止
# print("新一轮匹配,开始位置", cur)
for i in range(n): # 一次匹配长度
if s[i+cur] != p[i]:
# print(s[i+cur], p[i], '不匹配。查表位置:', i, i - table[i-1])
cur += max(i - table[i-1], 1) # 有了部分匹配表,我们不只是单纯的1位1位往右移,可以一次移动多位
break
else:
return cur
return -1
#部分匹配表
def partial_table(self, p):
'''''partial_table("ABCDABD") -> [0, 0, 0, 0, 1, 2, 0]'''
prefix = set()
table = [0]
for i in range(1, len(p)): # 从1开始进行前后缀比较
prefix.add(p[:i]) # 前缀每次累加就行
postfix = set()
for j in range(1, i+1): # i+1 因为i需要包括
postfix.add(p[j:i+1])
# print(prefix, postfix)
# print(prefix&postfix, len(prefix&postfix))
# table.append(len((sorted((prefix&postfix),key = len)or {''}).pop()))
if prefix&postfix:
table.append(max(map(len,prefix&postfix)))
else:
table.append(0)
return table
Java
Java的来自网络,有空理解下,实现table似乎原理不同
public class KMP {
public static int kmp(String str, String dest,int[] next){//str文本串 dest 模式串
for(int i = 0, j = 0; i < str.length(); i++){
while(j > 0 && str.charAt(i) != dest.charAt(j)){
j = next[j - 1];
}
if(str.charAt(i) == dest.charAt(j)){
j++;
}
if(j == dest.length()){
return i-j+1;
}
}
return 0;
}
public static int[] kmpnext(String dest){
int[] next = new int[dest.length()];
next[0] = 0;
for(int i = 1,j = 0; i < dest.length(); i++){
while(j > 0 && dest.charAt(j) != dest.charAt(i)){
j = next[j - 1];
}
if(dest.charAt(i) == dest.charAt(j)){
j++;
}
next[i] = j;
}
return next;
}
public static void main(String[] args){
String a = "aabaaac";
String b = "aabaaabaaac";
int[] next = kmpnext(a);
int res = kmp(b, a, next);
System.out.println("result:" + res);
for(int i = 0; i < next.length; i++){
System.out.println("table:" + next[i]);
}
System.out.println(next.length);
}
}