C++ 如何在C/C+；中检查字符串中的模式+；？

如果我有一个像

abcabc

这样的字符串，很明显，

abc

是一个模式。我想用c/c++找出模式

我不想实施。一个伪代码/算法就可以了

我怎么做

尝试查找以下内容：

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不要把它看作一个字符串，而是寻找一个句点。它是否是字符串并不重要。

使用弗洛伊德循环查找算法。这使用慢-快类比来寻找循环。维基百科提供的Python源代码：

def floyd(f, x0):
    # Main phase of algorithm: finding a repetition x_i = x_2i
    # The hare moves twice as quickly as the tortoise and
    # the distance between them increases by 1 at each step.
    # Eventually they will both be inside the cycle and then,
    # at some point, the distance between them will be
    # divisible by the period λ.
    tortoise = f(x0) # f(x0) is the element/node next to x0.
    hare = f(f(x0))
    while tortoise != hare:
        tortoise = f(tortoise)
        hare = f(f(hare))

    # At this point the tortoise position, ν, which is also equal
    # to the distance between hare and tortoise, is divisible by
    # the period λ. So hare moving in circle one step at a time, 
    # and tortoise (reset to x0) moving towards the circle, will 
    # intersect at the beginning of the circle. Because the 
    # distance between them is constant at 2ν, a multiple of λ,
    # they will agree as soon as the tortoise reaches index μ.

    # Find the position μ of first repetition.    
    mu = 0
    tortoise = x0
    while tortoise != hare:
        tortoise = f(tortoise)
        hare = f(hare)   # Hare and tortoise move at same speed
        mu += 1

    # Find the length of the shortest cycle starting from x_μ
    # The hare moves one step at a time while tortoise is still.
    # lam is incremented until λ is found.
    lam = 1
    hare = f(tortoise)
    while tortoise != hare:
        hare = f(hare)
        lam += 1

    return lam, mu

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此解决方案的时间复杂度为

O（λ，μ）

，辅助空间为

O（1）

一种方法是使用时间复杂度为O（p.length）的预计算算法，其中p为给定字符串，它计算一个查找表'PI'，该表包含与其相应前缀匹配的最长后缀（“a”、“ab”、“abc”和…）的长度

伪代码取自算法简介，CLRS。也， Linux对上述算法有相当好的理解

因此，p.length-PI[p.length]=k=最小重复图案的长度。记住，k将始终保持在[0，P.length]范围内

例如，“abcabc”=PI[0,0,0,1,2,3,4,5,6]。在这里，最小重复模式的长度为9-6=3。但k是否将字符串等分

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因此，如果p.length mod k==0？p[1..k]将是您的重复模式。

是否保证字符串具有模式？它是否具有有限长度？如果字符串确实包含重复模式，您可以使用更通用的模式查找算法来查找模式，请参阅。