搜索模式| 系列2—

给定一个文本txt [0..n-1]和一个模式pat [0..m-1]，写一个搜索函数search（char pat []，char txt []），在txt中打印所有出现的pat [] []。可以假设n> m。

例子：

Input:  txt[] = "THIS IS A TEST TEXT"
        pat[] = "TEST"
Output: Pattern found at index 10

Input:  txt[] =  "AABAACAADAABAABA"
        pat[] =  "AABA"
Output: Pattern found at index 0
        Pattern found at index 9
        Pattern found at index 12

模式搜索是计算机科学中的一个重要问题。当我们在记事本/ word文件或浏览器或数据库中搜索字符串时，使用模式搜索算法来显示搜索结果。

我们在前面的文章中已经讨论过Naive模式搜索算法。Naive算法的最坏情况复杂度是O（m（n-m + 1））。在最坏的情况下，KMP算法的时间复杂度是O（n）。

KMP（克努特莫里斯普拉特）模式搜索的Naive模式搜索算法并不在在这种情况下很好的工作：许多匹配字符，随后是不匹配的字符的情况下。以下是一些例子。

   txt[] = "AAAAAAAAAAAAAAAAAB"
   pat[] = "AAAAB"

   txt[] = "ABABABCABABABCABABABC"
   pat[] =  "ABABAC" (not a worst case, but a bad case for Naive)

KMP匹配算法使用该模式的退化性质（在模式中具有相同子模式出现不止一次的模式），并将最坏情况复杂度提高到O（n）。

KMP算法背后的基本思想是：每当我们检测到一个不匹配（在一些匹配之后），我们已经知道下一个窗口文本中的一些字符。我们利用这些信息避免匹配我们知道无论如何将匹配的字符。让我们考虑下面的例子来理解这一点。

Matching Overview
txt = "AAAAABAAABA"
pat = "AAAA"

我们对比txt和pat：
txt = "AAAAABAAABA"
pat = "AAAA"  初始位置
我们找到了一个和原始字符串相同的匹配。

下一步，我们比较txt中接下来的字符串和pat字符串
txt = "AAAAABAAABA"
pat =  "AAAA" [初始位置的下一个位置]
这就是KMP算法优于基本模式匹配算法的地方了。在第二次比较中，我们只比较了pattern的前4个'A'，然后我们决定当前是否匹配，我们已经知道前三个匹配，我们直接跳过前三个即可。

需要预处理吗？　
上述解释产生了一个重要问题，如何知道要跳过多少字符。要知道为此，我们预处理模式并准备一个整数数组。告诉我们要跳过的字符数。

预处理概述：

KMP算法对pat []进行预处理，并构造一个大小为m（与模式大小相同）的辅助lps []，用于匹配时跳过字符。
名称lps表示最长的正确前缀，也是后缀。。一个适当的前缀是前缀与整个字符串不容许。例如，“ABC”的前缀是“”，“A”，“AB”和“ABC”。正确的前缀是“”，“A”和“AB”。字符串的后缀是“”，“C”，“BC”和“ABC”。
对于其中i = 0到m-1的每个子模式pat [0..i]，lps [i]存储最大匹配正确前缀的长度，其也是子模式pat [0..i]的后缀。

 lps[i] = the longest proper prefix of pat[..i]
              which ..i].

注意： lps [i]也可以定义为最长的前缀，也是正确的后缀。我们需要在一个地方使用正确的，以确保整个子字符串不被考虑。

Examples of lps[] construction:
For the pattern “AAAA”,
lps[] , , , ]

For the pattern “ABCDE”,
lps[] , , , , ]

For the pattern “AABAACAABAA”,
lps[] , , , , , , , , , , ]

For the pattern “AAACAAAAAC”,
lps[] , , , , , , , , , ] 

For the pattern “AAABAAA”,
lps[] , , , , , , ]

搜索算法：
与Naive算法不同的是，我们将模式依次匹配一个，然后比较每次匹配中中的所有字符，我们使用来自lps []的值来决定下一个要匹配的字符。这个想法是不匹配我们知道无论如何将匹配的字符。

如何使用lps []来决定下一个职位（或知道要跳过的字符数）？

我们开始与j = 0的pat [j]与当前文本窗口的字符进行比较。
我们保持匹配字符txt [i]和pat [j]，并在pat [j]和txt [i]保持匹配的同时不断增加i和j 。
当我们看到不匹配的时候
- 我们知道，字符pat [0..j-1]与txt [i-j + 1 ... i-1]匹配（注意，j从0开始并且只有在匹配时才增加）。
- 我们也知道（从上面的定义中）lps [j-1]是pat [0 ... j-1]的字符数，它们都是正确的前缀和后缀。
- 从上面两点我们可以得出结论，我们不需要将这些lps [j-1]字符与txt [ij ... i-1]进行匹配，因为我们知道这些字符无论如何将匹配。让我们考虑上面的例子来理解这一点。

txt[] = "AAAAABAAABA"
pat[] = "AAAA"
lps[] = {0, 1, 2, 3} 

i = 0, j = 0
txt[] = "AAAAABAAABA"
pat[] = "AAAA"
txt[i] and pat[j[ match, do i++, j++

i = 1, j = 1
txt[] = "AAAAABAAABA"
pat[] = "AAAA"
txt[i] and pat[j[ match, do i++, j++

i = 2, j = 2
txt[] = "AAAAABAAABA"
pat[] = "AAAA"
pat[i] and pat[j[ match, do i++, j++

i = 3, j = 3
txt[] = "AAAAABAAABA"
pat[] = "AAAA"
txt[i] and pat[j[ match, do i++, j++

i = 4, j = 4
Since j == M, print pattern found and resset j,
j = lps[j-1] = lps[3] = 3

Here unlike Naive algorithm, we do not match first three
characters of this window. Value of lps[j-1] (in above
step) gave us index of next character to match.
i = 4, j = 3
txt[] = "AAAAABAAABA"
pat[] =  "AAAA"
txt[i] and pat[j[ match, do i++, j++

i = 5, j = 4
Since j == M, print pattern found and reset j,
j = lps[j-1] = lps[3] = 3

Again unlike Naive algorithm, we do not match first three
characters of this window. Value of lps[j-1] (in above
step) gave us index of next character to match.
i = 5, j = 3
txt[] = "AAAAABAAABA"
pat[] =   "AAAA"
txt[i] and pat[j] do NOT match and j > 0, change only j
j = lps[j-1] = lps[2] = 2

i = 5, j = 2
txt[] = "AAAAABAAABA"
pat[] =    "AAAA"
txt[i] and pat[j] do NOT match and j > 0, change only j
j = lps[j-1] = lps[1] = 1 

i = 5, j = 1
txt[] = "AAAAABAAABA"
pat[] =     "AAAA"
txt[i] and pat[j] do NOT match and j > 0, change only j
j = lps[j-1] = lps[0] = 0

i = 5, j = 0
txt[] = "AAAAABAAABA"
pat[] =      "AAAA"
txt[i] and pat[j] do NOT match and j is 0, we do i++.

i = 6, j = 0
txt[] = "AAAAABAAABA"
pat[] =       "AAAA"
txt[i] and pat[j] match, do i++ and j++

i = 7, j = 1
txt[] = "AAAAABAAABA"
pat[] =       "AAAA"
txt[i] and pat[j] match, do i++ and j++

　　下面是代码实现：

// C++ program for implementation of KMP pattern searching
// algorithm
#include<bits/stdc++.h>

void computeLPSArray(char *pat, int M, int *lps);

// Prints occurrences of txt[] in pat[]
void KMPSearch(char *pat, char *txt)
{
    int M = strlen(pat);
    int N = strlen(txt);

    // create lps[] that will hold the longest prefix suffix
    // values for pattern
    int lps[M];

    // Preprocess the pattern (calculate lps[] array)
    computeLPSArray(pat, M, lps);

    ;  // index for txt[]
    ;  // index for pat[]
    while (i < N)
    {
        if (pat[j] == txt[i])
        {
            j++;
            i++;
        }

        if (j == M)
        {
            printf("Found pattern at index %d n", i-j);
            j = lps[j-];
        }

        // mismatch after j matches
        else if (i < N && pat[j] != txt[i])
        {
            // Do not match lps[0..lps[j-1]] characters,
            // they will match anyway
            )
                j = lps[j-];
            else
                i = i+;
        }
    }
}

// Fills lps[] for given patttern pat[0..M-1]
void computeLPSArray(char *pat, int M, int *lps)
{
    // length of the previous longest prefix suffix
    ;

    lps[] = ; // lps[0] is always 0

    // the loop calculates lps[i] for i = 1 to M-1
    ;
    while (i < M)
    {
        if (pat[i] == pat[len])
        {
            len++;
            lps[i] = len;
            i++;
        }
        else // (pat[i] != pat[len])
        {
            // This is tricky. Consider the example.
            // AAACAAAA and i = 7. The idea is similar
            // to search step.
            )
            {
                len = lps[len-];

                // Also, note that we do not increment
                // i here
            }
            else // if (len == 0)
            {
                lps[i] = ;
                i++;
            }
        }
    }
}

// Driver program to test above function
int main()
{
    char *txt = "ABABDABACDABABCABAB";
    char *pat = "ABABCABAB";
    KMPSearch(pat, txt);
    ;
}

输出：

Found pattern at index

预处理算法：
在预处理部分，我们计算lps []中的值。为此，我们跟踪前一个索引的最长前缀后缀值（我们使用len变量用于此目的）的长度。我们将lps [0]和len初始化为0.如果pat [len]和pat [i]匹配，我们将len加1，并将增加的值赋给lps [i]。如果pat [i]和pat [len]不匹配，len不为0，我们将len更新为lps [len-1]。有关详细信息，请参阅下面的代码中的computeLPSArray（）。

预处理的插图（或构建lps []）

pat[] = "AAACAAAA"

len = , i  = .
lps[] , we move
to i = 

len = , i  = .
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = , lps[] = , i = 

len = , i  = .
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = , lps[] = , i = 

len = , i  = .
Since pat[len] and pat[i] ,
] = lps[] = 

len = , i  = .
Since pat[len] and pat[i] ,
len = lps[len-] = lps[] = 

len = , i  = .
Since pat[len] and pat[i] ,
Set lps[] =  and i = .

len = , i  = .
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = , lps[] = , i = 

len = , i  = .
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = , lps[] = , i = 

len = , i  = .
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = , lps[] = , i = 

len = , i  = .
Since pat[len] and pat[i] ,
] = lps[] = 

len = , i  = .
Since pat[len] and pat[i] match, do len++,
store it in lps[i] and do i++.
len = , lps[] = , i = 

We stop here as we have constructed the whole lps[].

ok，如果有问题，随时提问

巴特西

搜索模式| 系列2——KMP算法

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