背景
后缀数组：
后缀数组是给定字符串的所有后缀的排序数组。
让给定的字符串为" banana"。
0 banana 5 a
1 anana Sort the Suffixes 3 ana
2 nana ----------------> 1 anana
3 ana alphabetically 0 banana
4 na 4 na
5 a 2 nana" banana"的后缀数组：
后缀[] = {5, 3, 1, 0, 4, 2}
我们已经讨论过后缀数组及其O(nLogn)构造.
构建后缀数组后, 我们可以使用它来有效地搜索文本中的模式。例如, 我们可以使用Binary Search查找模式(讨论了相同的完整代码这里)
LCP阵列
讨论了基于二进制搜索的解决方案
这里
花费O(m * Logn)时间, 其中m是要搜索的模式的长度, n是文本的长度。借助LCP阵列, 我们可以搜索O(m + Log n)时间中的模式。例如, 如果我们的任务是在"香蕉"中搜索" ana", 则m = 3, n = 5。
LCP数组是大小为n的数组(如后缀数组)。值lcp [i]表示后缀[i]和后缀[i + 1]互不相同的后缀的最长公共前缀的长度。后缀[n-1]未定义, 因为其后没有后缀。
txt[0..n-1] = "banana"
suffix[] = {5, 3, 1, 0, 4, 2|
lcp[] = {1, 3, 0, 0, 2, 0}
Suffixes represented by suffix array in order are:
{"a", "ana", "anana", "banana", "na", "nana"}
lcp[0] = Longest Common Prefix of "a" and "ana" = 1
lcp[1] = Longest Common Prefix of "ana" and "anana" = 3
lcp[2] = Longest Common Prefix of "anana" and "banana" = 0
lcp[3] = Longest Common Prefix of "banana" and "na" = 0
lcp[4] = Longest Common Prefix of "na" and "nana" = 2
lcp[5] = Longest Common Prefix of "nana" and None = 0如何构造LCP阵列？
LCP阵列的构建有两种方式：
1)计算LCP数组作为后缀数组的副产品(Manber和Myers算法)
2)使用已经构造的后缀数组以计算LCP值。 (Kasai算法)。
有一些算法可以在O(n)时间内构造后缀数组, 因此我们总是可以在O(n)时间内构造LCP数组。但是在下面的实现中, 将讨论O(n Log n)算法。
kasai的算法
本文讨论了kasai的算法。该算法根据后缀数组和O(n)时间的输入文本构造LCP数组。这个想法基于以下事实：
令以txt [i [开头的后缀lcp为k。如果k大于0, 则从txt [i + 1]开始的后缀的lcp至少为k-1。原因是, 字符的相对顺序保持不变。如果我们从两个后缀中删除第一个字符, 我们知道至少将匹配k个字符。例如, 对于子字符串" ana", lcp为3, 因此对于字符串" na", lcp至少为2。这个为证明。
以下是Kasai算法的C ++实现。
// C++ program for building LCP array for given text
#include <bits/stdc++.h>
using namespace std;
// Structure to store information of a suffix
struct suffix
{
int index; // To store original index
int rank[2]; // To store ranks and next rank pair
};
// A comparison function used by sort() to compare two suffixes
// Compares two pairs, returns 1 if first pair is smaller
int cmp( struct suffix a, struct suffix b)
{
return (a.rank[0] == b.rank[0])? (a.rank[1] < b.rank[1] ?1: 0):
(a.rank[0] < b.rank[0] ?1: 0);
}
// This is the main function that takes a string 'txt' of size n as an
// argument, builds and return the suffix array for the given string
vector< int > buildSuffixArray(string txt, int n)
{
// A structure to store suffixes and their indexes
struct suffix suffixes[n];
// Store suffixes and their indexes in an array of structures.
// The structure is needed to sort the suffixes alphabatically
// and maintain their old indexes while sorting
for ( int i = 0; i < n; i++)
{
suffixes[i].index = i;
suffixes[i].rank[0] = txt[i] - 'a' ;
suffixes[i].rank[1] = ((i+1) < n)? (txt[i + 1] - 'a' ): -1;
}
// Sort the suffixes using the comparison function
// defined above.
sort(suffixes, suffixes+n, cmp);
// At his point, all suffixes are sorted according to first
// 2 characters. Let us sort suffixes according to first 4
// characters, then first 8 and so on
int ind[n]; // This array is needed to get the index in suffixes[]
// from original index. This mapping is needed to get
// next suffix.
for ( int k = 4; k < 2*n; k = k*2)
{
// Assigning rank and index values to first suffix
int rank = 0;
int prev_rank = suffixes[0].rank[0];
suffixes[0].rank[0] = rank;
ind[suffixes[0].index] = 0;
// Assigning rank to suffixes
for ( int i = 1; i < n; i++)
{
// If first rank and next ranks are same as that of previous
// suffix in array, assign the same new rank to this suffix
if (suffixes[i].rank[0] == prev_rank &&
suffixes[i].rank[1] == suffixes[i-1].rank[1])
{
prev_rank = suffixes[i].rank[0];
suffixes[i].rank[0] = rank;
}
else // Otherwise increment rank and assign
{
prev_rank = suffixes[i].rank[0];
suffixes[i].rank[0] = ++rank;
}
ind[suffixes[i].index] = i;
}
// Assign next rank to every suffix
for ( int i = 0; i < n; i++)
{
int nextindex = suffixes[i].index + k/2;
suffixes[i].rank[1] = (nextindex < n)?
suffixes[ind[nextindex]].rank[0]: -1;
}
// Sort the suffixes according to first k characters
sort(suffixes, suffixes+n, cmp);
}
// Store indexes of all sorted suffixes in the suffix array
vector< int >suffixArr;
for ( int i = 0; i < n; i++)
suffixArr.push_back(suffixes[i].index);
// Return the suffix array
return suffixArr;
}
/* To construct and return LCP */
vector< int > kasai(string txt, vector< int > suffixArr)
{
int n = suffixArr.size();
// To store LCP array
vector< int > lcp(n, 0);
// An auxiliary array to store inverse of suffix array
// elements. For example if suffixArr[0] is 5, the
// invSuff[5] would store 0. This is used to get next
// suffix string from suffix array.
vector< int > invSuff(n, 0);
// Fill values in invSuff[]
for ( int i=0; i < n; i++)
invSuff[suffixArr[i]] = i;
// Initialize length of previous LCP
int k = 0;
// Process all suffixes one by one starting from
// first suffix in txt[]
for ( int i=0; i<n; i++)
{
/* If the current suffix is at n-1, then we don’t
have next substring to consider. So lcp is not
defined for this substring, we put zero. */
if (invSuff[i] == n-1)
{
k = 0;
continue ;
}
/* j contains index of the next substring to
be considered to compare with the present
substring, i.e., next string in suffix array */
int j = suffixArr[invSuff[i]+1];
// Directly start matching from k'th index as
// at-least k-1 characters will match
while (i+k<n && j+k<n && txt[i+k]==txt[j+k])
k++;
lcp[invSuff[i]] = k; // lcp for the present suffix.
// Deleting the starting character from the string.
if (k>0)
k--;
}
// return the constructed lcp array
return lcp;
}
// Utility function to print an array
void printArr(vector< int >arr, int n)
{
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
cout << endl;
}
// Driver program
int main()
{
string str = "banana" ;
vector< int >suffixArr = buildSuffixArray(str, str.length());
int n = suffixArr.size();
cout << "Suffix Array : \n" ;
printArr(suffixArr, n);
vector< int >lcp = kasai(str, suffixArr);
cout << "\nLCP Array : \n" ;
printArr(lcp, n);
return 0;
}输出如下：
Suffix Array :
5 3 1 0 4 2
LCP Array :
1 3 0 0 2 0插图：
txt[] = "banana", suffix[] = {5, 3, 1, 0, 4, 2|
Suffix array represents
{"a", "ana", "anana", "banana", "na", "nana"}
Inverse Suffix Array would be
invSuff[] = {3, 2, 5, 1, 4, 0}LCP值按以下顺序评估
我们首先计算文本中第一个后缀的LCP, 即"香蕉"。我们需要后缀数组中的下一个后缀来计算LCP(请记住, lcp [i]被定义为后缀[i]和后缀[i + 1]的最长公共前缀)。要在suffixArr []中找到下一个后缀, 我们使用SuffInv []。下一个后缀是" na"。由于" banana"和" na"之间没有通用前缀, 因此" banana"的LCP值为0, 并且在后缀数组中位于索引3, 因此我们填充lcp [3]为0。
接下来, 我们计算第二个后缀的LCP安娜娜"。后缀数组中" anana"的下一个后缀是" banana"。由于没有公共前缀, 因此" anana"的LCP的值为0, 并且在后缀数组的索引2处, 因此我们填充lcp [2]为0。
接下来, 我们计算第三个后缀的LCP娜娜"。由于没有下一个后缀, 因此未定义" nana"的LCP值。我们填写lcp [5]为0。
文本中的下一个后缀是" ana"。下一个后缀"安娜后缀数组中的""是" anana"。由于存在长度为3的公共前缀, 因此" ana"的LCP值为3。我们填充lcp [1]如3。
现在, 我们在文本中的下一个后缀lcpna"。这是Kasai算法使用的技巧, 因为先前的LCP值为3, 所以LCP值必须至少为2。由于" na"后没有字符, 因此LCP的最终值为2。lcp [4]如2。
文字的下一个后缀是"一种"。 LCP值必须至少为1, 因为先前的值为2。由于在" a"之后没有字符, 因此LCP的最终值为1。lcp [0]如1。
我们很快将在LCP阵列的帮助下讨论搜索的实现, 以及LCP阵列如何帮助将时间复杂度降低到O(m + Log n)。
参考文献：
http://web.stanford.edu/class/cs97si/suffix-array.pdf
http://www.mi.fu-berlin.de/wiki/pub/ABI/RnaSeqP4/suffix-array.pdf
http://codeforces.com/blog/entry/12796
本文作者：Prakhar Agrawal。如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请发表评论。
![从字法上最小长度N的排列，使得对于正好为K个索引，a[i] a[i]+1](https://www.lsbin.com/wp-content/themes/begin%20lts/img/loading.png)
