给定字符串str ，任务是找到str及其每个后缀的相似度之和。
字符串A和B的相似性是最长前缀共同的长度到两个字符串，即“AABC”和“AAB”的相似性是3并且“QWER”和“ABC”的值为0。
例子：
Input: str = “ababa”
Output: 9
The suffixes of str are “ababa”, “baba”, “aba”, “ba” and “a”. The similarities of these strings with the original string “ababa” are 5, 0, 3, 0 & 1 respectively.
Thus, the answer is 5 + 0 + 3 + 0 + 1 = 9.
Input: str = “aaabaab”
Output: 13
方法：使用Z算法计算Z数组–对于字符串str [0..n-1]，Z数组的长度与字符串的长度相同。 Z数组的元素Z [i]存储从str [i]开始的最长子串的长度，该字符串也是str [0..n-1]的前缀。 Z数组的第一个条目是字符串的长度。
现在，对Z数组的所有元素求和，以获得所需的相似度总和。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
#include
#include
using namespace std;
// Function to calculate the Z-array for the given string
void getZarr(string str, int n, int Z[])
{
int L, R, k;
// [L, R] make a window which matches with prefix of s
L = R = 0;
for (int i = 1; i < n; ++i) {
// if i>R nothing matches so we will calculate.
// Z[i] using naive way.
if (i > R) {
L = R = i;
// R-L = 0 in starting, so it will start
// checking from 0'th index. For example,
// for "ababab" and i = 1, the value of R
// remains 0 and Z[i] becomes 0. For string
// "aaaaaa" and i = 1, Z[i] and R become 5
while (R < n && str[R - L] == str[R])
R++;
Z[i] = R - L;
R--;
}
else {
// k = i-L so k corresponds to number which
// matches in [L, R] interval.
k = i - L;
// if Z[k] is less than remaining interval
// then Z[i] will be equal to Z[k].
// For example, str = "ababab", i = 3, R = 5
// and L = 2
if (Z[k] < R - i + 1)
Z[i] = Z[k];
// For example str = "aaaaaa" and i = 2, R is 5,
// L is 0
else {
// else start from R and check manually
L = i;
while (R < n && str[R - L] == str[R])
R++;
Z[i] = R - L;
R--;
}
}
}
}
// Function to return the similarity sum
int sumSimilarities(string s, int n)
{
int Z[n] = { 0 };
// Compute the Z-array for the given string
getZarr(s, n, Z);
int total = n;
// Summation of the Z-values
for (int i = 1; i < n; i++)
total += Z[i];
return total;
}
// Driver code
int main()
{
string s = "ababa";
int n = s.length();
cout << sumSimilarities(s, n);
return 0;
} Java
// Java implementation of the above approach
public class GFG{
// Function to calculate the Z-array for the given string
static void getZarr(String str, int n, int Z[])
{
int L, R, k;
// [L, R] make a window which matches with prefix of s
L = R = 0;
for (int i = 1; i < n; ++i) {
// if i>R nothing matches so we will calculate.
// Z[i] using naive way.
if (i > R) {
L = R = i;
// R-L = 0 in starting, so it will start
// checking from 0'th index. For example,
// for "ababab" and i = 1, the value of R
// remains 0 and Z[i] becomes 0. For string
// "aaaaaa" and i = 1, Z[i] and R become 5
while (R < n && str.charAt(R - L) == str.charAt(R))
R++;
Z[i] = R - L;
R--;
}
else {
// k = i-L so k corresponds to number which
// matches in [L, R] interval.
k = i - L;
// if Z[k] is less than remaining interval
// then Z[i] will be equal to Z[k].
// For example, str = "ababab", i = 3, R = 5
// and L = 2
if (Z[k] < R - i + 1)
Z[i] = Z[k];
// For example str = "aaaaaa" and i = 2, R is 5,
// L is 0
else {
// else start from R and check manually
L = i;
while (R < n && str.charAt(R - L) == str.charAt(R))
R++;
Z[i] = R - L;
R--;
}
}
}
}
// Function to return the similarity sum
static int sumSimilarities(String s, int n)
{
int Z[] = new int[n] ;
// Compute the Z-array for the given string
getZarr(s, n, Z);
int total = n;
// Summation of the Z-values
for (int i = 1; i < n; i++)
total += Z[i];
return total;
}
// Driver code
public static void main(String []args)
{
String s = "ababa";
int n = s.length();
System.out.println(sumSimilarities(s, n));
}
// This code is contributed by Ryuga
}Python3
# Python3 implementation of the approach
def getZarr(s, n, Z):
L, R, k = 0, 0, 0
# [L, R] make a window which matches
# with prefix of s
for i in range(n):
# if i>R nothing matches so we will
# calculate Z[i] using naive way.
if i > R:
L, R = i, i
'''
R-L = 0 in starting, so it will start
checking from 0'th index. For example,
for "ababab" and i = 1, the value of R
remains 0 and Z[i] becomes 0. For string
"aaaaaa" and i = 1, Z[i] and R become 5
'''
while R < n and s[R - L] == s[R]:
R += 1
Z[i] = R - L
R -= 1
else:
# k = i-L so k corresponds to number
# which matches in [L, R] interval.
k = i - L
# if Z[k] is less than remaining interval
# then Z[i] will be equal to Z[k].
# For example, str = "ababab", i = 3, R = 5
# and L = 2
if Z[k] < R - i + 1:
Z[i] = Z[k]
else:
L = i
while R < n and s[R - L] == s[R]:
R += 1
Z[i] = R - L
R -= 1
def sumSimilarities(s, n):
Z = [0 for i in range(n)]
# Compute the Z-array for the
# given string
getZarr(s, n, Z)
total = n
# summation of the Z-values
for i in range(n):
total += Z[i]
return total
# Driver Code
s = "ababa"
n = len(s)
print(sumSimilarities(s, n))
# This code is contributed
# by Mohit kumar 29C#
//C# implementation of the above approach
using System;
public class GFG{
// Function to calculate the Z-array for the given string
static void getZarr(string str, int n, int []Z)
{
int L, R, k;
// [L, R] make a window which matches with prefix of s
L = R = 0;
for (int i = 1; i < n; ++i) {
// if i>R nothing matches so we will calculate.
// Z[i] using naive way.
if (i > R) {
L = R = i;
// R-L = 0 in starting, so it will start
// checking from 0'th index. For example,
// for "ababab" and i = 1, the value of R
// remains 0 and Z[i] becomes 0. For string
// "aaaaaa" and i = 1, Z[i] and R become 5
while (R < n && str[R - L] == str[R])
R++;
Z[i] = R - L;
R--;
}
else {
// k = i-L so k corresponds to number which
// matches in [L, R] interval.
k = i - L;
// if Z[k] is less than remaining interval
// then Z[i] will be equal to Z[k].
// For example, str = "ababab", i = 3, R = 5
// and L = 2
if (Z[k] < R - i + 1)
Z[i] = Z[k];
// For example str = "aaaaaa" and i = 2, R is 5,
// L is 0
else {
// else start from R and check manually
L = i;
while (R < n && str[R - L] == str[R])
R++;
Z[i] = R - L;
R--;
}
}
}
}
// Function to return the similarity sum
static int sumSimilarities(string s, int n)
{
int []Z = new int[n] ;
// Compute the Z-array for the given string
getZarr(s, n, Z);
int total = n;
// Summation of the Z-values
for (int i = 1; i < n; i++)
total += Z[i];
return total;
}
// Driver code
static public void Main (){
string s = "ababa";
int n = s.Length;
Console.WriteLine(sumSimilarities(s, n));
}
// This code is contributed by ajit.
}输出：
9