算法4第6章后缀数组讲解
最长重复子字符串/后缀数组
给定字符串怎样找到它的最长重复子字符串，如字符串atobeornottobe，最长重复子字符串是tobe
使用一般的方法速度很慢，时间复杂度是O(N^2),而使用后缀数组可以巧妙而高效的解决此问题。
首先找出字符的所有后缀字符串组成后缀字符串数组，对数组进行排序，然后遍历数组，最长重复子字符串就在相邻的两个后缀字符串中
如字符串tobeornottobe，后缀字符串是atobeornottobe,tobeornottobe,obeornottobe,beornottobe,eornottobe,ornottobe,rnottobe，nottobe，ottobe，ttobe,tobe,obe,be,e
排序后就是
atobeornottobe
be
beornottobe
e
eornottobe
nottobe
obeornottobe
ornottobe
ottobe
rnottobe
ttobe
tobe
tobeornottobe
该算法在一般情况下时间复杂度是O(NlogN),代码如下
public class LongestRepeatedSubstring {
// Do not instantiate.
private LongestRepeatedSubstring() { }
/**
* Returns the longest common string of the two specified strings.
*
* @param s one string
* @param t the other string
* @return the longest common string that appears as a substring
*/
/**
* Returns the longest repeated substring of the specified string.
*
* @param text the string
* @return the longest repeated substring that appears in {@code text};
* the empty string if no such string
*/
public static String lrs(String text) {
int n = text.length();
SuffixArray sa = new SuffixArray(text);
String lrs = "";
for (int i = 1; i < n; i++) {
int length = sa.lcp(i);
if (length > lrs.length()) {
// lrs = sa.select(i).substring(0, length);
lrs = text.substring(sa.index(i), sa.index(i) + length);
}
}
return lrs;
}
/**
* Unit tests the {@code lrs()} method.
*
* @param args the command-line arguments
*/
public static void main(String[] args) {
String text = StdIn.readAll().replaceAll("\\s+", " ");
StdOut.println("'" + lrs(text) + "'");
}
}
public class SuffixArray {
private Suffix[] suffixes;
/**
* Initializes a suffix array for the given {@code text} string.
* @param text the input string
*/
public SuffixArray(String text) {
int n = text.length();
this.suffixes = new Suffix[n];
for (int i = 0; i < n; i++)
suffixes[i] = new Suffix(text, i);
Arrays.sort(suffixes);
}
private static class Suffix implements Comparable
private final String text;
private final int index;
private Suffix(String text, int index) {
this.text = text;
this.index = index;
}
private int length() {
return text.length() - index;
}
private char charAt(int i) {
return text.charAt(index + i);
}
public int compareTo(Suffix that) {
if (this == that) return 0; // optimization
int n = Math.min(this.length(), that.length());
for (int i = 0; i < n; i++) {
if (this.charAt(i) < that.charAt(i)) return -1;
if (this.charAt(i) > that.charAt(i)) return +1;
}
return this.length() - that.length();
}
public String toString() {
return text.substring(index);
}
}
/**
* Returns the length of the input string.
* @return the length of the input string
*/
public int length() {
return suffixes.length;
}
/**
* Returns the index into the original string of the ith smallest suffix.
* That is, {@code text.substring(sa.index(i))} is the ith smallest suffix.
* @param i an integer between 0 and n-1
* @return the index into the original string of the ith smallest suffix
* @throws java.lang.IllegalArgumentException unless {@code 0 <= i < n}
*/
public int index(int i) {
if (i < 0 || i >= suffixes.length) throw new IllegalArgumentException();
return suffixes[i].index;
}
/**
* Returns the length of the longest common prefix of the ith
* smallest suffix and the i-1st smallest suffix.
* @param i an integer between 1 and n-1
* @return the length of the longest common prefix of the ith
* smallest suffix and the i-1st smallest suffix.
* @throws java.lang.IllegalArgumentException unless {@code 1 <= i < n}
*/
public int lcp(int i) {
if (i < 1 || i >= suffixes.length) throw new IllegalArgumentException();
return lcpSuffix(suffixes[i], suffixes[i-1]);
}
// longest common prefix of s and t
private static int lcpSuffix(Suffix s, Suffix t) {
int n = Math.min(s.length(), t.length());
for (int i = 0; i < n; i++) {
if (s.charAt(i) != t.charAt(i)) return i;
}
return n;
}
/**
* Returns the ith smallest suffix as a string.
* @param i the index
* @return the i smallest suffix as a string
* @throws java.lang.IllegalArgumentException unless {@code 0 <= i < n}
*/
public String select(int i) {
if (i < 0 || i >= suffixes.length) throw new IllegalArgumentException();
return suffixes[i].toString();
}
/**
* Returns the number of suffixes strictly less than the {@code query} string.
* We note that {@code rank(select(i))} equals {@code i} for each {@code i}
* between 0 and n-1.
* @param query the query string
* @return the number of suffixes strictly less than {@code query}
*/
public int rank(String query) {
int lo = 0, hi = suffixes.length - 1;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
int cmp = compare(query, suffixes[mid]);
if (cmp < 0) hi = mid - 1;
else if (cmp > 0) lo = mid + 1;
else return mid;
}
return lo;
}
// compare query string to suffix
private static int compare(String query, Suffix suffix) {
int n = Math.min(query.length(), suffix.length());
for (int i = 0; i < n; i++) {
if (query.charAt(i) < suffix.charAt(i)) return -1;
if (query.charAt(i) > suffix.charAt(i)) return +1;
}
return query.length() - suffix.length();
}
/**
* Unit tests the {@code SuffixArray} data type.
*
* @param args the command-line arguments
*/
public static void main(String[] args) {
String s = StdIn.readAll().replaceAll("\\s+", " ").trim();
SuffixArray suffix = new SuffixArray(s);
// StdOut.println("rank(" + args[0] + ") = " + suffix.rank(args[0]));
StdOut.println(" i ind lcp rnk select");
StdOut.println("---------------------------");
for (int i = 0; i < s.length(); i++) {
int index = suffix.index(i);
String ith = "\"" + s.substring(index, Math.min(index + 50, s.length())) + "\"";
assert s.substring(index).equals(suffix.select(i));
int rank = suffix.rank(s.substring(index));
if (i == 0) {
StdOut.printf("%3d %3d %3s %3d %s\n", i, index, "-", rank, ith);
}
else {
int lcp = suffix.lcp(i);
StdOut.printf("%3d %3d %3d %3d %s\n", i, index, lcp, rank, ith);
}
}
}
}