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String Matching

[编程题]String Matching

热度指数：11047 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 64M，其他语言128M
算法知识视频讲解

Finding all occurrences of a pattern in a text is a problem that arises frequently in text-editing programs. Typically,the text is a document being edited,and the pattern searched for is a particular word supplied by the user. We assume that the text is an array T[1..n] of length n and that the pattern is an array P[1..m] of length m<=n.We further assume that the elements of P and T are all alphabets(∑={a,b...,z}).The character arrays P and T are often called strings of characters. We say that pattern P occurs with shift s in the text T if 0<=s<=n and T[s+1..s+m] = P[1..m](that is if T[s+j]=P[j],for 1<=j<=m). If P occurs with shift s in T,then we call s a valid shift;otherwise,we calls a invalid shift. Your task is to calculate the number of vald shifts for the given text T and p attern P.

输入描述:

   For each case, there are two strings T and P on a line,separated by a single space.You may assume both the length of T and P will not exceed 10^6.

输出描述:

    You should output a number on a separate line,which indicates the number of valid shifts for the given text T and pattern P.

示例1

输入

abababab abab

输出

算法知识视频讲解

Python3

chong_0428

def getNext(t):
    next = [0] * (len(t)+1)
    t = list(t)
    i, k = 0, -1
    next[0] = -1
    while i < len(t):
        if k == -1&nbs***bsp;t[i] == t[k]:
            k+=1
            i+=1
            next[i] = k
        else:
            k = next[k]
    return next

def kmp(s, t):
    i, j = 0, 0
    res =0
    next = getNext(t)
    s = list(s)
    t = list(t)
    while i < len(s) and j < int(len(t)):
        if(j == -1&nbs***bsp;s[i] == t[j]):
            i+=1
            j+=1
        else:
            j = next[j]

        if j == len(t):
            res+=1
            j = next[j]
            #return i - j
    return res
   

# while True:
#     try:
s, t = input().split()
res = kmp(s, t)
print(res)
    # except:
    #     break

编辑于 2024-03-19 18:13:11 回复(0)