给定一个数组和一个整数K。我们需要找到长度为K的每个分段的最大值，该分段中没有重复项。
例子：
Input : a[] = {1, 2, 2, 3, 3},
K = 3.
Output : 1 3 2
For segment (1, 2, 2), Maximum = 1.
For segment (2, 2, 3), Maximum = 3.
For segment (2, 3, 3), Maximum = 2.
Input : a[] = {3, 3, 3, 4, 4, 2},
K = 4.
Output : 4 Nothing 3一个简单的解决方案是运行两个循环。对于每个子数组，找到所有不同的元素并打印最大的唯一元素。
一个有效的解决方案是使用滑动窗口技术。我们在每个窗口中维护两个结构。
1)哈希表，用于存储当前窗口中所有元素的计数。
2)自平衡BST(使用C++ STL中的set和Java的TreeSet来实现)。这个想法是快速找到最大元素并更新最大元素。
我们处理第一个K-1元素，并将它们的计数存储在哈希表中。我们还将独特的元素存储在集合中。现在我们逐个处理每个窗口的最后一个元素。如果当前元素是唯一的，则将其添加到集合中。我们还增加了它的数量。处理完最后一个元素后，我们从集合中打印最大值。在开始下一次迭代之前，我们删除上一个窗口的第一个元素。
C++
// C++ code to calculate maximum unique
// element of every segment of array
#include
using namespace std;
void find_max(int A[], int N, int K)
{
// Storing counts of first K-1 elements
// Also storing distinct elements.
map Count;
for (int i = 0; i < K - 1; i++)
Count[A[i]]++;
set Myset;
for (auto x : Count)
if (x.second == 1)
Myset.insert(x.first);
// Before every iteration of this loop,
// we maintain that K-1 elements of current
// window are processed.
for (int i = K - 1; i < N; i++) {
// Process K-th element of current window
Count[A[i]]++;
if (Count[A[i]] == 1)
Myset.insert(A[i]);
else
Myset.erase(A[i]);
// If there are no distinct
// elements in current window
if (Myset.size() == 0)
printf("Nothing\n");
// Set is ordered and last element
// of set gives us maximum element.
else
printf("%d\n", *Myset.rbegin());
// Remove first element of current
// window before next iteration.
int x = A[i - K + 1];
Count[x]--;
if (Count[x] == 1)
Myset.insert(x);
if (Count[x] == 0)
Myset.erase(x);
}
}
// Driver code
int main()
{
int a[] = { 1, 2, 2, 3, 3 };
int n = sizeof(a) / sizeof(a[0]);
int k = 3;
find_max(a, n, k);
return 0;
} Java
// Java code to calculate maximum unique
// element of every segment of array
import java.io.*;
import java.util.*;
class GFG {
static void find_max(int[] A, int N, int K)
{
// Storing counts of first K-1 elements
// Also storing distinct elements.
HashMap Count = new HashMap<>();
for (int i = 0; i < K - 1; i++)
if (Count.containsKey(A[i]))
Count.put(A[i], 1 + Count.get(A[i]));
else
Count.put(A[i], 1);
TreeSet Myset = new TreeSet();
for (Map.Entry x : Count.entrySet()) {
if (Integer.parseInt(String.valueOf(x.getValue())) == 1)
Myset.add(Integer.parseInt(String.valueOf(x.getKey())));
}
// Before every iteration of this loop,
// we maintain that K-1 elements of current
// window are processed.
for (int i = K - 1; i < N; i++) {
// Process K-th element of current window
if (Count.containsKey(A[i]))
Count.put(A[i], 1 + Count.get(A[i]));
else
Count.put(A[i], 1);
if (Integer.parseInt(String.valueOf(Count.get(A[i]))) == 1)
Myset.add(A[i]);
else
Myset.remove(A[i]);
// If there are no distinct
// elements in current window
if (Myset.size() == 0)
System.out.println("Nothing");
// Set is ordered and last element
// of set gives us maximum element.
else
System.out.println(Myset.last());
// Remove first element of current
// window before next iteration.
int x = A[i - K + 1];
Count.put(x, Count.get(x) - 1);
if (Integer.parseInt(String.valueOf(Count.get(x))) == 1)
Myset.add(x);
if (Integer.parseInt(String.valueOf(Count.get(x))) == 0)
Myset.remove(x);
}
}
// Driver code
public static void main(String args[])
{
int[] a = { 1, 2, 2, 3, 3 };
int n = a.length;
int k = 3;
find_max(a, n, k);
}
}
// This code is contributed by rachana soma Python3
# Python3 code to calculate maximum unique
# element of every segment of array
def find_max(A, N, K):
# Storing counts of first K-1 elements
# Also storing distinct elements.
Count = dict()
for i in range(K - 1):
Count[A[i]] = Count.get(A[i], 0) + 1
Myset = dict()
for x in Count:
if (Count[x] == 1):
Myset[x] = 1
# Before every iteration of this loop,
# we maintain that K-1 elements of current
# window are processed.
for i in range(K - 1, N):
# Process K-th element of current window
Count[A[i]] = Count.get(A[i], 0) + 1
if (Count[A[i]] == 1):
Myset[A[i]] = 1
else:
del Myset[A[i]]
# If there are no distinct
# elements in current window
if (len(Myset) == 0):
print("Nothing")
# Set is ordered and last element
# of set gives us maximum element.
else:
maxm = -10**9
for i in Myset:
maxm = max(i, maxm)
print(maxm)
# Remove first element of current
# window before next iteration.
x = A[i - K + 1]
if x in Count.keys():
Count[x] -= 1
if (Count[x] == 1):
Myset[x] = 1
if (Count[x] == 0):
del Myset[x]
# Driver code
a = [1, 2, 2, 3, 3 ]
n = len(a)
k = 3
find_max(a, n, k)
# This code is contributed
# by mohit kumarC#
using System;
using System.Collections.Generic;
public class GFG
{
static void find_max(int[] A, int N, int K)
{
// Storing counts of first K-1 elements
// Also storing distinct elements.
Dictionary count = new Dictionary();
for (int i = 0; i < K - 1; i++)
{
if(count.ContainsKey(A[i]))
{
count[A[i]]++;
}
else
{
count.Add(A[i], 1);
}
}
HashSet Myset = new HashSet();
foreach(KeyValuePair x in count)
{
if(x.Value == 1)
{
Myset.Add(x.Key);
}
}
// Before every iteration of this loop,
// we maintain that K-1 elements of current
// window are processed.
for (int i = K - 1; i < N; i++)
{
// Process K-th element of current window
if (count.ContainsKey(A[i]))
{
count[A[i]]++;
}
else
{
count.Add(A[i], 1);
}
if(count[A[i]] == 1)
{
Myset.Add(A[i]);
}
else
{
Myset.Remove(A[i]);
}
// If there are no distinct
// elements in current window
if (Myset.Count == 0)
Console.Write("Nothing\n");
// Set is ordered and last element
// of set gives us maximum element.
else
{
List myset = new List(Myset);
Console.WriteLine(myset[myset.Count - 1]);
}
// Remove first element of current
// window before next iteration.
int x = A[i - K + 1];
count[x]--;
if(count[x] == 1)
{
Myset.Add(x);
}
if(count[x] == 0)
{
Myset.Remove(x);
}
}
}
// Driver code
static public void Main ()
{
int[] a = { 1, 2, 2, 3, 3 };
int n=a.Length;
int k = 3;
find_max(a, n, k);
}
}
// This code is contributed by rag2127 输出：
1
3
2时间复杂度： O(N Log K)