给定一个只包含小写英文字母的字符数组arr[] ，任务是打印子数组的最大长度，使得子数组的第一个和最后一个元素相同。
例子：
Input: arr[] = {‘g’, ‘e’, ‘e’, ‘k’, ‘s’}
Output: 2
{‘e’, ‘e’} is the maximum length sub-array satisfying the given condition.
Input: arr[] = {‘a’, ‘b’, ‘c’, ‘d’, ‘a’}
Output: 5
{‘a’, ‘b’, ‘c’, ‘d’, ‘a’} is the required sub-array
方法：对于数组ch 的每个元素，存储它的第一次和最后一次出现。那么以相同元素ch开始和结束的子数组的最大长度将是lastOccurrence(ch) – firstOccurrence(ch) + 1 。所有元素中该值的最大值就是要求的答案。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Class that represents a single element
// of the given array and stores it's first
// and the last occurrence in the array
class Element
{
public:
int firstOcc, lastOcc;
Element();
void updateOccurence(int);
};
Element::Element()
{
firstOcc = lastOcc = -1;
}
// Function to update the occurrence
// of a particular character in the array
void Element::updateOccurence(int index)
{
// If first occurrence is set to something
// other than -1 then it doesn't need updating
if (firstOcc == -1)
firstOcc = index;
// Last occurrence will be updated everytime
// the character appears in the array
lastOcc = index;
}
// Function to return the maximum length of the
// sub-array that starts and ends with the same element
int maxLenSubArr(string arr, int n)
{
Element elements[26];
for (int i = 0; i < n; i++)
{
int ch = arr[i] - 'a';
// Update current character's occurrence
elements[ch].updateOccurence(i);
}
int maxLen = 0;
for (int i = 0; i < 26; i++)
{
// Length of the longest sub-array that starts
// and ends with the same element
int len = elements[i].lastOcc -
elements[i].firstOcc + 1;
maxLen = max(maxLen, len);
}
// Return the maximum length of
// the required sub-array
return maxLen;
}
// Driver Code
int main()
{
string arr = "geeks";
int n = arr.length();
cout << maxLenSubArr(arr, n) << endl;
return 0;
}
// This code is contributed by
// sanjeev2552 Java
// Java implementation of the approach
// Class that represents a single element
// of the given array and stores it's first
// and the last occurrence in the array
class Element {
int firstOcc, lastOcc;
public Element()
{
firstOcc = lastOcc = -1;
}
// Function to update the occurrence
// of a particular character in the array
public void updateOccurrence(int index)
{
// If first occurrence is set to something
// other than -1 then it doesn't need updating
if (firstOcc == -1)
firstOcc = index;
// Last occurrence will be updated everytime
// the character appears in the array
lastOcc = index;
}
}
class GFG {
// Function to return the maximum length of the
// sub-array that starts and ends with the same element
public static int maxLenSubArr(char arr[], int n)
{
Element elements[] = new Element[26];
for (int i = 0; i < n; i++) {
int ch = arr[i] - 'a';
// Initialize the current character
// if haven't already
if (elements[ch] == null)
elements[ch] = new Element();
// Update current character's occurrence
elements[ch].updateOccurrence(i);
}
int maxLen = 0;
for (int i = 0; i < 26; i++) {
// If current character appears in the given array
if (elements[i] != null) {
// Length of the longest sub-array that starts
// and ends with the same element
int len = elements[i].lastOcc - elements[i].firstOcc + 1;
maxLen = Math.max(maxLen, len);
}
}
// Return the maximum length of
// the required sub-array
return maxLen;
}
// Driver code
public static void main(String[] args)
{
char arr[] = { 'g', 'e', 'e', 'k', 's' };
int n = arr.length;
System.out.print(maxLenSubArr(arr, n));
}
}Python3
# Python3 implementation of the approach
# Class that represents a single element
# of the given array and stores it's first
# and the last occurrence in the array
class Element:
def __init__(self):
self.firstOcc = -1
self.lastOcc = -1
# Function to update the occurrence
# of a particular character in the array
def updateOccurrence(self, index):
# If first occurrence is set to
# something other than -1 then it
# doesn't need updating
if self.firstOcc == -1:
self.firstOcc = index
# Last occurrence will be updated
# everytime the character appears
# in the array
self.lastOcc = index
# Function to return the maximum length
# of the sub-array that starts and ends
# with the same element
def maxLenSubArr(arr, n):
elements = [None] * 26
for i in range(0, n):
ch = ord(arr[i]) - ord('a')
# Initialize the current character
# if haven't already
if elements[ch] == None:
elements[ch] = Element()
# Update current character's occurrence
elements[ch].updateOccurrence(i)
maxLen = 0
for i in range(0, 26):
# If current character appears in
# the given array
if elements[i] != None:
# Length of the longest sub-array that
# starts and ends with the same element
length = (elements[i].lastOcc -
elements[i].firstOcc + 1)
maxLen = max(maxLen, length)
# Return the maximum length of
# the required sub-array
return maxLen
# Driver code
if __name__ == "__main__":
arr = ['g', 'e', 'e', 'k', 's']
n = len(arr)
print(maxLenSubArr(arr, n))
# This code is contributed by Rituraj JainC#
// C# implementation of the above approach
using System;
// Class that represents a single element
// of the given array and stores it's first
// and the last occurrence in the array
public class Element
{
public int firstOcc, lastOcc;
public Element()
{
firstOcc = lastOcc = -1;
}
// Function to update the occurrence
// of a particular character in the array
public void updateOccurrence(int index)
{
// If first occurrence is set to something
// other than -1 then it doesn't need updating
if (firstOcc == -1)
firstOcc = index;
// Last occurrence will be updated everytime
// the character appears in the array
lastOcc = index;
}
}
class GFG
{
// Function to return the maximum
// length of the sub-array that
// starts and ends with the same element
public static int maxLenSubArr(char []arr, int n)
{
Element []elements = new Element[26];
for (int i = 0; i < n; i++)
{
int ch = arr[i] - 'a';
// Initialize the current character
// if haven't already
if (elements[ch] == null)
elements[ch] = new Element();
// Update current character's occurrence
elements[ch].updateOccurrence(i);
}
int maxLen = 0;
for (int i = 0; i < 26; i++)
{
// If current character appears
// in the given array
if (elements[i] != null)
{
// Length of the longest sub-array that starts
// and ends with the same element
int len = elements[i].lastOcc - elements[i].firstOcc + 1;
maxLen = Math.Max(maxLen, len);
}
}
// Return the maximum length of
// the required sub-array
return maxLen;
}
// Driver code
public static void Main()
{
char []arr = { 'g', 'e', 'e', 'k', 's' };
int n = arr.Length;
Console.WriteLine(maxLenSubArr(arr, n));
}
}
// This code is contributed by Ryuga输出：
2时间复杂度：O(N)
辅助空间： O(N)
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