根据后缀树得出后缀数组
后缀数组（suffix array）是字符串S的所有后缀按照一定顺序（通常是字典顺序）排列后得出的数组，数组中每个元素的值等于该后缀的起始下标。要根据后缀树得出后缀数组，只需遍历所有的叶子结点（按照从左到右的顺序依次访问每个叶子结点），因为每一个叶子结点对应一个后缀，它的起始下标=字符串S的长度-叶子结点的pathlen值。
代码：
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
/**
*
* Derive the suffix array from the suffix tree
* (The suffix tree is built with ukk algorithm)
* @author ljs
* 2011-07-13
*
*/
public class SuffixTree2Array {
private class SuffixNode {
private StringBuilder sb;
private List<SuffixNode> children = new LinkedList<SuffixNode>();
private SuffixNode link;
private int start;
private int end;
private int pathlen;
public SuffixNode(StringBuilder sb,int start,int end,int pathlen){
this.sb = sb;
this.start = start;
this.end = end;
this.pathlen = pathlen;
}
public SuffixNode(StringBuilder sb){
this.sb = sb;
this.start = -1;
this.end = -1;
this.pathlen = 0;
}
public int getLength(){
if(start == -1) return 0;
else return end - start + 1;
}
public String getString(){
if(start != -1){
return this.sb.substring(start,end+1);
}else{
return "";
}
}
public boolean isRoot(){
return start == -1;
}
public String getCoordinate(){
return "[" + start+".." + end + "/" + this.pathlen + "]";
}
public String toString(){
return getString() + "(" + getCoordinate()
+ ",link:" + ((this.link==null)?"N/A":this.link.getCoordinate())
+ ",children:" + children.size() +")";
}
}
private class State{
private SuffixNode u; //parent(v)
//private SuffixNode w;
private SuffixNode v;
//private int k; //the global index of text starting from 0 to text.length()
//private boolean finished;
}
private SuffixNode root;
private StringBuilder sb = new StringBuilder();
//build a suffix-tree for a string of text
public void buildSuffixTree(String text) throws Exception{
int m = text.length();
if(m==0)
return;
if(root==null){
root = new SuffixNode(sb);
root.link = root; //link to itself
}
List<SuffixNode> leaves = new ArrayList<SuffixNode>();
//add first node
sb.append(text.charAt(0));
SuffixNode node = new SuffixNode(sb,0,0,1);
leaves.add(node);
root.children.add(node);
int j_star = 0; //j_{i-1}
SuffixNode u = root;
SuffixNode v = root;
for(int i=1;i<=m-1;i++){
//do phase i
sb.append(text.charAt(i));
//step 1: do implicit extensions
for(SuffixNode leafnode:leaves){
leafnode.end++;
leafnode.pathlen++;
}
//step 2: do explicit extensions until rule #3 is applied
State state = new State();
//for the first explicit extension, we reuse the last phase's u and do slowscan
//also note: suffix link doesn't span two phases.
int j=j_star+1;
SuffixNode s = u;
int k = s.pathlen + j;
state.u = s;
state.v = s;
SuffixNode newleaf = slowscan(state,s,k);
if(newleaf == null){
//if rule #3 is applied, then we can terminate this phase
j_star = j - 1;
//Note: no need to update state.v because it is not going to be used
//at the next phase
u = state.u;
continue;
}else{
j_star = j;
leaves.add(newleaf);
u = state.u;
v = state.v;
}
j++;
//for other explicit extensions, we start with fast scan.
for(;j<=i;j++){
s = u.link;
int uvLen=v.pathlen - u.pathlen;
if(u.isRoot() && !v.isRoot()){
uvLen--;
}
//starting with index k of the text
k = s.pathlen + j;
//init state
state.u = s;
state.v = s; //if uvLen = 0
//execute fast scan
newleaf = fastscan(state,s,uvLen,k);
//establish the suffix link with v
v.link = state.v;
if(newleaf == null){
//if rule #3 is applied, then we can terminate this phase
j_star = j - 1;
u = state.u;
break;
}else{
j_star = j;
leaves.add(newleaf);
u = state.u;
v = state.v;
}
}
}
}
//slow scan from currNode until state.v is found
//return the new leaf if a new one is created right after v;
//return null otherwise (i.e. when rule #3 is applied)
private SuffixNode slowscan(State state,SuffixNode currNode,int k){
SuffixNode newleaf = null;
boolean done = false;
int keyLen = sb.length() - k;
for(int i=0;i<currNode.children.size();i++){
SuffixNode child = currNode.children.get(i);
//use min(child.key.length, key.length)
int childKeyLen = child.getLength();
int len = childKeyLen<keyLen?childKeyLen:keyLen;
int delta = 0;
for(;delta<len;delta++){
if(sb.charAt(k+delta) != sb.charAt(child.start+delta)){
break;
}
}
if(delta==0){//this child doesn't match any character with the new key
//order keys by lexi-order
if(sb.charAt(k) < sb.charAt(child.start)){
//e.g. child="e" (currNode="abc")
// abc abc
// / \ =========> / | \
// e f insert "c" c e f
int pathlen = sb.length() - k + currNode.pathlen;
SuffixNode node = new SuffixNode(sb,k,sb.length()-1,pathlen);
currNode.children.add(i,node);
//state.u = currNode; //currNode is already registered as state.u, so commented out
state.v = currNode;
newleaf = node;
done = true;
break;
}else{ //key.charAt(0)>child.key.charAt(0)
//don't forget to add the largest new key after iterating all children
continue;
}
}else{//current child's key partially matches with the new key
if(delta==len){
if(keyLen==childKeyLen){
//e.g. child="ab"
// ab ab
// / \ =========> / \
// e f insert "ab" e f
//terminate this phase (implicit tree with rule #3)
state.u = child;
state.v = currNode;
}else if(keyLen>childKeyLen){
//TODO: still need an example to test this condition
//e.g. child="ab"
// ab ab
// / \ ==========> / | \
// e f insert "abc" c e f
//recursion
state.u = child;
state.v = child;
k += childKeyLen;
//state.k = k;
newleaf = slowscan(state,child,k);
}
else{ //keyLen<childKeyLen
//e.g. child="abc"
// abc abc
// / \ =========> / \
// e f insert "ab" e f
//
//terminate this phase (implicit tree with rule #3)
//state.u = currNode;
state.v = currNode;
}
}else{//0<delta<len
//e.g. child="abc"
// abc ab
// / \ ==========> / \
// e f insert "abd" c d
// / \
// e f
//insert the new node: ab
int nodepathlen = child.pathlen
- (child.getLength()-delta);
SuffixNode node = new SuffixNode(sb,
child.start,child.start + delta - 1,nodepathlen);
node.children = new LinkedList<SuffixNode>();
int leafpathlen = (sb.length() - (k + delta)) + nodepathlen;
SuffixNode leaf = new SuffixNode(sb,
k+delta,sb.length()-1,leafpathlen);
//update child node: c
child.start += delta;
if(sb.charAt(k+delta)<sb.charAt(child.start)){
node.children.add(leaf);
node.children.add(child);
}else{
node.children.add(child);
node.children.add(leaf);
}
//update parent
currNode.children.set(i, node);
//state.u = currNode; //currNode is already registered as state.u, so commented out
state.v = node;
newleaf = leaf;
}
done = true;
break;
}
}
if(!done){
int pathlen = sb.length() - k + currNode.pathlen;
SuffixNode node = new SuffixNode(sb,k,sb.length()-1,pathlen);
currNode.children.add(node);
//state.u = currNode; //currNode is already registered as state.u, so commented out
state.v = currNode;
newleaf = node;
}
return newleaf;
}
//fast scan until state.v is found;
//return the new leaf if a new one is created right after v;
//return null otherwise (i.e. when rule #3 is applied)
private SuffixNode fastscan(State state,SuffixNode currNode,int uvLen,int k){
if(uvLen==0){
//state.u = currNode; //currNode is already registered as state.u, so commented out
//continue with slow scan
return slowscan(state,currNode,k);
}
SuffixNode newleaf = null;
boolean done = false;
for(int i=0;i<currNode.children.size();i++){
SuffixNode child = currNode.children.get(i);
if(sb.charAt(child.start) == sb.charAt(k)){
int len = child.getLength();
if(uvLen==len){
//then we find v
//uvLen = 0;
state.u = child;
//state.v = child;
k += len;
//state.k = k;
//continue with slow scan
newleaf = slowscan(state,child,k);
}else if(uvLen<len){
//we know v must be an internal node; branching and cut child short
//e.g. child="abc",uvLen = 2
// abc ab
// / \ ================> / \
// e f suffix part: "abd" c d
// / \
// e f
//insert the new node: ab; child is now c
int nodepathlen = child.pathlen
- (child.getLength()-uvLen);
SuffixNode node = new SuffixNode(sb,
child.start,child.start + uvLen - 1,nodepathlen);
node.children = new LinkedList<SuffixNode>();
int leafpathlen = (sb.length() - (k + uvLen)) + nodepathlen;
SuffixNode leaf = new SuffixNode(sb,
k+uvLen,sb.length()-1,leafpathlen);
//update child node: c
child.start += uvLen;
if(sb.charAt(k+uvLen)<sb.charAt(child.start)){
node.children.add(leaf);
node.children.add(child);
}else{
node.children.add(child);
node.children.add(leaf);
}
//update parent
currNode.children.set(i, node);
//uvLen = 0;
//state.u = currNode; //currNode is already registered as state.u, so commented out
state.v = node;
newleaf = leaf;
}else{//uvLen>len
//e.g. child="abc", uvLen = 4
// abc
// / \ ================>
// e f suffix part: "abcde"
//
//
//jump to next node
uvLen -= len;
state.u = child;
//state.v = child;
k += len;
//state.k = k;
newleaf = fastscan(state,child,uvLen,k);
}
done = true;
break;
}
}
if(!done){
//TODO: still need an example to test this condition
//add a leaf under the currNode
int pathlen = sb.length() - k + currNode.pathlen;
SuffixNode node = new SuffixNode(sb,k,sb.length()-1,pathlen);
currNode.children.add(node);
//state.u = currNode; //currNode is already registered as state.u, so commented out
state.v = currNode;
newleaf = node;
}
return newleaf;
}
public void toSuffixArray(int[] S){
toSuffixArray(root,S,0);
}
public int toSuffixArray(SuffixNode currNode,int[] S,int j){
for(int i=0;i<currNode.children.size();i++){
SuffixNode child = currNode.children.get(i);
if(child.children.size()==0){
//leaf
S[j++] = sb.length() - child.pathlen;
}else{
j=toSuffixArray(child,S,j);
}
}
return j;
}
//for test purpose only
public void printTree(){
System.out.format("The suffix tree for S = %s is: %n",this.sb);
this.print(0, this.root);
}
private void print(int level, SuffixNode node){
for (int i = 0; i < level; i++) {
System.out.format(" ");
}
System.out.format("|");
for (int i = 0; i < level; i++) {
System.out.format("-");
}
System.out.format("%s(%d..%d/%d)%n", node.getString(),node.start,node.end,node.pathlen);
//System.out.format("(%d,%d)%n", node.start,node.end);
for (SuffixNode child : node.children) {
print(level + 1, child);
}
}
public static void main(String[] args) throws Exception {
//test suffix-tree
System.out.println("****************************");
String text = "yabbadabbado{1}quot;; //the last char must be unique!
SuffixTree2Array stree = new SuffixTree2Array();
stree.buildSuffixTree(text);
//stree.printTree();
int[] S = new int[text.length()];
stree.toSuffixArray(S);
System.out.format("Suffix array for \"%s\" is:%n",text);
for(int i:S){
System.out.format("%d ", i);
}
System.out.println();
System.out.println("****************************");
text = "banana{1}quot;;
stree = new SuffixTree2Array();
stree.buildSuffixTree(text);
//stree.printTree();
S = new int[text.length()];
stree.toSuffixArray(S);
System.out.format("Suffix array for \"%s\" is:%n",text);
for(int i:S){
System.out.format("%d ", i);
}
System.out.println();
System.out.println("****************************");
text = "GACCCACCACC{1}quot;;
stree = new SuffixTree2Array();
stree.buildSuffixTree(text);
//stree.printTree();
S = new int[text.length()];
stree.toSuffixArray(S);
System.out.format("Suffix array for \"%s\" is:%n",text);
for(int i:S){
System.out.format("%d ", i);
}
System.out.println();
System.out.println("****************************");
text = "mississippi{1}quot;;
stree = new SuffixTree2Array();
stree.buildSuffixTree(text);
//stree.printTree();
S = new int[text.length()];
stree.toSuffixArray(S);
System.out.format("Suffix array for \"%s\" is:%n",text);
for(int i:S){
System.out.format("%d ", i);
}
System.out.println();
}
}
测试：
****************************Suffix array for "yabbadabbado$" is:
12 1 6 4 9 3 8 2 7 5 10 11 0
****************************
Suffix array for "banana$" is:
6 5 3 1 0 4 2
****************************
Suffix array for "GACCCACCACC$" is:
11 8 5 1 10 7 4 9 6 3 2 0
****************************
Suffix array for "mississippi$" is:
11 10 7 4 1 0 9 8 6 3 5 2