倍增算法求解字符串的后缀数组

    今天完成了使用倍增算法求解字符串的后缀数组的程序。倍增算法的主要思想是：

对每个字符开始的长度为2k的子字符串进行排序，求出排名，即Rank值。k从0开始，每次加1，当2k大于n以后，每个字符开始的长度为2k的子字符串便相当于所有的后缀。并且这些子字符串都一定已经比较出大小，即Rank值中没有相同的值，那么此时的Rank值就是最后的结果。每一次排序都利用上次长度为2k-1的字符串的Rank值，那么长度为2k的字符串就可以用两个长度为2k-1的字符串的排名作为关键字表示，然后进行基数排序，便得出了长度为2k的字符串的Rank值。

因为我们知道有Rank[i] = j 等价于 SA[j] = i；故可以利用这一点求出相应字符串的后缀数组。其时间复杂度是O(nlogn)。

 

 

 #include <iostream>#include <string>#include <cmath>using namespace std;void select(int *arrays,string str){for(unsigned i = 0; i < str.length();i++){switch(str.at(i)){case '$':arrays[i] = 1;break;case 'b':arrays[i] = 2;break;case 'e':arrays[i] = 3;break;case 'n':arrays[i] = 4;break;case 'o':arrays[i] = 5;break;case 'r':arrays[i] = 6;break;case 't':arrays[i] = 7;break;default:exit(0);}}}const int MAX = 20;typedef struct radixArray{int x;int y;int pos;}Radix_Array;Radix_Array RadixArr[MAX];typedef struct bnode{Radix_Array data;struct bnode *next;}node;typedef struct Bucket{int no;node *next;}bucket;bucket bcket[14];void RadixSort(Radix_Array RadixArr[],int *Rank,int n){for(int i = 0; i < 14;i++){bcket[i].no = 0;bcket[i].next = NULL;}for(int i = 0; i < n;i++){node *tmp = new node;tmp->data.x = RadixArr[i].x;tmp->data.y = RadixArr[i].y;tmp->data.pos = RadixArr[i].pos;tmp->next = NULL;if(bcket[tmp->data.x - 1].next == NULL){bcket[tmp->data.x - 1].next = tmp;}else{node *tm = bcket[tmp->data.x - 1].next;node *per = new node;while(tm->next != NULL && tm->data.y <= tmp->data.y){per = tm;tm = tm->next;}if(tm->next == NULL){//插到桶链表的第一个节点if(tm == bcket[tmp->data.x - 1].next && tm->data.y > tmp->data.y){bcket[tmp->data.x - 1].next = tmp;tmp->next = tm;}//插入到桶链表的倒数第一个节点之前else if (tm->data.y > tmp->data.y){per->next = tmp;tmp->next = tm;}else //插入到桶链表的最后一个节点tm->next = tmp;}else //插入到桶的中间某个节点{ per->next = tmp;tmp->next = tm;}}}for(int j = 0; j < 14; j++){node *tm = bcket[j].next;while(tm != NULL){cout<<"("<<tm->data.x<<","<<tm->data.y<<","<<tm->data.pos<<") ";tm = tm->next;}}cout<<endl;//Rank 初始值。。注意结构体中pos的应用，，，int v = 1;for(int i = 0; i < 14; i++){node *tm = bcket[i].next;while(tm != NULL){Rank[tm->data.pos] = v;node *pre = tm;if(tm->next != NULL){if(tm->next->data.y != tm->data.y){v++;Rank[tm->next->data.pos] =v;}elseRank[tm->next->data.pos] = v;}tm = tm->next;}v++;}for(int i = 0; i < n; i++)cout<<Rank[i]<<" ";cout<<endl;}int main(int argc,char **argv){//string str = "aabaaaab";string str = "tobeornottobe$";int *arrays = new int[str.length()];int *Rank = new int[str.length()];int *SA = new int[str.length()+1];select(arrays,str);for(unsigned i = 0; i < str.length();i++)Rank[i] = arrays[i];int h = 1;int exp = (int)pow((double)2,(double)(h-1));while(exp <= str.length()){for(unsigned i = 0; i < str.length(); i++){RadixArr[i].x = Rank[i];if(i+exp<str.length())RadixArr[i].y = Rank[i+exp];elseRadixArr[i].y = 0;RadixArr[i].pos = i;}for(unsigned i = 0; i < str.length(); i++)cout<<"( "<< RadixArr[i].x<<","<<RadixArr[i].y<<","<<RadixArr[i].pos<<")"<< " ";cout<<endl<<endl;cout<<endl;RadixSort(RadixArr,Rank,str.length());h++;exp = (int)pow((double)2,(double)(h-1));}for(unsigned i = 0; i < str.length();i++)SA[Rank[i]] = i+1;//输出SA arraycout<<"字符串 str = tobeornottobe$;后缀数组是："<<endl;for(unsigned i = 1; i < str.length()+1;i++)cout<<SA[i]<<" ";system("pause");}

该程序实现了后缀数组的O(nlogn)时间复杂度的求解。唯一的问题就是空间复杂度比paper中提到的要高。