找出最长的子串，它是前缀、后缀，也存在于串内|集合 2
原文:https://www . geesforgeks . org/find-最长的子字符串-前缀后缀兼现字符串集-2/
给定弦弦。任务是找到最长的子字符串，它是给定字符串 str 的前缀、后缀和子字符串。如果不存在这样的字符串，则打印 -1 。 例:
输入:str = " geeksisfurgecksinplatformgekers " 输出:gekers 输入:str = " fixeprefixsuffix " 输出: fix
注:本文第一集附此处为。 方法: 实现采用 BIT 和 Z 算法。从本和本了解更多。
- 首先我们使用 Z 算法计算 Z 数组。
- 从索引 z[i]将位数组中的值更新 1。
- 使用 pref 函数查询所需的最大长度子字符串。
- 如果 len 为 0，则从给定的字符串中不可能出现这样的子字符串。
下面是实现方式:
C++
// C++ program to find that
// substring which is its
// suffix prefix and also
// found somewhere in between
#include <bits/stdc++.h>
using namespace std;
// Z-algorithm function
vector<int> z_function(string s)
{
int n = s.size();
vector<int> z(n);
for (int i = 1, l = 0,
r = 0;
i < n;
i++) {
if (i <= r)
z[i] = min(r - i + 1,
z[i - l]);
while (i + z[i] < n
&& s[z[i]] == s[i + z[i]])
z[i]++;
if (i + z[i] - 1 > r)
l = i, r = i + z[i] - 1;
}
return z;
}
int n, len = 0;
// BIT array
int bit[1000005];
string s;
vector<int> z;
map<int, int> m;
// bit update function which
// updates values from index
// "idx" to last by value "val"
void update(int idx, int val)
{
if (idx == 0)
return;
while (idx <= n) {
bit[idx] += val;
idx += (idx & -idx);
}
}
// Query function in bit
int pref(int idx)
{
int ans = 0;
while (idx > 0) {
ans += bit[idx];
idx -= (idx & -idx);
}
return ans;
}
// Driver Code
int main()
{
s = "geeksisforgeeksinplatformgeeks";
n = s.size();
// Making the z array
z = z_function(s);
// update in the bit array from
// index z[i] by increment of 1
for (int i = 1; i < n; i++) {
update(z[i], 1);
}
for (int i = n - 1; i > 1; i--) {
// if the value in z[i] is
// not equal to (n-i) then no
// need to move further
if (z[i] != (n - i))
continue;
// queryng for the maximum
// length substring from
// bit array
if (pref(n) - pref(z[i] - 1) >= 2) {
len = max(len, z[i]);
}
}
if (!len)
cout << "-1";
else
cout << s.substr(0, len);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find that
// substring which is its
// suffix prefix and also
// found somewhere in between
class GFG
{
// Z-algorithm function
static int[] z_function(char []s)
{
int n = s.length;
int []z = new int[n];
for (int i = 1, l = 0, r = 0;
i < n; i++)
{
if (i <= r)
z[i] = Math.min(r - i + 1,
z[i - l]);
while (i + z[i] < n &&
s[z[i]] == s[i + z[i]])
z[i]++;
if (i + z[i] - 1 > r)
{
l = i; r = i + z[i] - 1;
}
}
return z;
}
static int n, len = 0;
// BIT array
static int []bit = new int[1000005];
static String s;
static int[] z;
// bit update function which
// updates values from index
// "idx" to last by value "val"
static void update(int idx, int val)
{
if (idx == 0)
return;
while (idx <= n)
{
bit[idx] += val;
idx += (idx & -idx);
}
}
// Query function in bit
static int pref(int idx)
{
int ans = 0;
while (idx > 0)
{
ans += bit[idx];
idx -= (idx & -idx);
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
s = "geeksisforgeeksinplatformgeeks";
z = new int[s.length()];
n = s.length();
// Making the z array
z = z_function(s.toCharArray());
// update in the bit array from
// index z[i] by increment of 1
for (int i = 1; i < n; i++)
{
update(z[i], 1);
}
for (int i = n - 1; i > 1; i--)
{
// if the value in z[i] is
// not equal to (n-i) then no
// need to move further
if (z[i] != (n - i))
continue;
// queryng for the maximum
// length substring from
// bit array
if (pref(n) - pref(z[i] - 1) >= 2)
{
len = Math.max(len, z[i]);
}
}
if (len == 0)
System.out.println("-1");
else
System.out.println(s.substring(0, len));
}
}
// This code is contributed
// by PrinciRaj1992
Python 3
# Python3 program to find that
# substring which is its
# suffix prefix and also
# found somewhere in between
# Z-algorithm function
def z_function(s):
global z, n
n = len(s)
z = [0] * n
l, r = 0, 0
for i in range(1, n):
if i <= r:
z[i] = min(r - i + 1, z[i - 1])
while (i + z[i] < n and s[z[i]] == s[i + z[i]]):
z[i] += 1
if (i + z[i] - 1 > r):
l = i
r = i + z[i] - 1
return z
# bit update function which
# updates values from index
# "idx" to last by value "val"
def update(idx, val):
global bit
if idx == 0:
return
while idx <= n:
bit[idx] += val
idx += (idx & -idx)
# Query function in bit
def pref(idx):
global bit
ans = 0
while idx > 0:
ans += bit[idx]
idx -= (idx & -idx)
return ans
# Driver Code
if __name__ == "__main__":
n = 0
length = 0
# BIT array
bit = [0] * 1000005
z = []
m = dict()
s = "geeksisforgeeksinplatformgeeks"
# Making the z array
z = z_function(s)
# update in the bit array from
# index z[i] by increment of 1
for i in range(1, n):
update(z[i], 1)
for i in range(n - 1, 1, -1):
# if the value in z[i] is
# not equal to (n-i) then no
# need to move further
if z[i] != n - i:
continue
# queryng for the maximum
# length substring from
# bit array
if (pref(n) - pref(z[i] - 1)) >= 2:
length = max(length, z[i])
if not length:
print(-1)
else:
print(s[:length])
# This code is contributed by
# sanjeev2552
C
// C# program to find that
// substring which is its
// suffix prefix and also
// found somewhere in between
using System;
class GFG
{
// Z-algorithm function
static int[] z_function(char []s)
{
int n = s.Length;
int []z = new int[n];
for (int i = 1, l = 0, r = 0;
i < n; i++)
{
if (i <= r)
z[i] = Math.Min(r - i + 1,
z[i - l]);
while (i + z[i] < n &&
s[z[i]] == s[i + z[i]])
z[i]++;
if (i + z[i] - 1 > r)
{
l = i; r = i + z[i] - 1;
}
}
return z;
}
static int n, len = 0;
// BIT array
static int []bit = new int[1000005];
static String s;
static int[] z;
// bit update function which
// updates values from index
// "idx" to last by value "val"
static void update(int idx, int val)
{
if (idx == 0)
return;
while (idx <= n)
{
bit[idx] += val;
idx += (idx & -idx);
}
}
// Query function in bit
static int pref(int idx)
{
int ans = 0;
while (idx > 0)
{
ans += bit[idx];
idx -= (idx & -idx);
}
return ans;
}
// Driver Code
public static void Main(String[] args)
{
s = "geeksisforgeeksinplatformgeeks";
z = new int[s.Length];
n = s.Length;
// Making the z array
z = z_function(s.ToCharArray());
// update in the bit array from
// index z[i] by increment of 1
for (int i = 1; i < n; i++)
{
update(z[i], 1);
}
for (int i = n - 1; i > 1; i--)
{
// if the value in z[i] is
// not equal to (n-i) then no
// need to move further
if (z[i] != (n - i))
continue;
// queryng for the maximum
// length substring from
// bit array
if (pref(n) - pref(z[i] - 1) >= 2)
{
len = Math.Max(len, z[i]);
}
}
if (len == 0)
Console.WriteLine("-1");
else
Console.WriteLine(s.Substring(0, len));
}
}
// This code is contributed by PrinciRaj1992
Output:
geeks
时间复杂度: O(N)
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