给定N个具有其容量的盒子，该数量表示它可以在其上面容纳的盒子总数。只要每个盒子上方的盒子总数小于或等于其容量，就可以将盒子彼此堆叠。通过使用所有框找到可堆叠的最小数量。
例子：
Input: arr[] = {0, 0, 1, 1, 2}
Output: 2
First stack (top to bottom): 0 1 2
Second stack (top to bottom): 0 1
Input: arr[] = {1, 1, 4, 4}
Output: 1
All the boxes can be put on a single stack.
方法：让我们来一张地图，其中map [X]表示我们可以使用的容量为X的盒子数量。让我们一一构建堆栈。最初，堆栈的大小为0，然后，我们贪婪地遍历整个地图，并选择尽可能多的当前容量的框。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the count
// of minimum stacks
int countPiles(int n, int a[])
{
// Keep track of occurrence
// of each capacity
map occ;
// Fill the occurrence map
for (int i = 0; i < n; i++)
occ[a[i]]++;
// Number of piles is 0 initially
int pile = 0;
// Traverse occurrences in increasing
// order of capacities.
while (occ.size()) {
// Adding a new pile
pile++;
int size = 0;
unordered_set toRemove;
// Traverse all piles in increasing
// order of capacities
for (auto tm : occ) {
int mx = tm.first;
int ct = tm.second;
// Number of boxes of capacity mx
// that can be added to current pile
int use = min(ct, mx - size + 1);
// Update the occurrence
occ[mx] -= use;
// Update the size of the pile
size += use;
if (occ[mx] == 0)
toRemove.insert(mx);
}
// Remove capacities that are
// no longer available
for (auto tm : toRemove)
occ.erase(tm);
}
return pile;
}
// Driver code
int main()
{
int a[] = { 0, 0, 1, 1, 2 };
int n = sizeof(a) / sizeof(a[0]);
cout << countPiles(n, a);
return 0;
} Java
// Java implementation of the approach
import java.util.HashMap;
import java.util.HashSet;
class GFG
{
// Function to return the count
// of minimum stacks
static int countPiles(int n, int[] a)
{
// Keep track of occurrence
// of each capacity
HashMap occ = new HashMap<>();
// Fill the occurrence map
for (int i = 0; i < n; i++)
occ.put(a[i], occ.get(a[i]) == null ? 1 :
occ.get(a[i]) + 1);
// Number of piles is 0 initially
int pile = 0;
// Traverse occurrences in increasing
// order of capacities.
while (!occ.isEmpty())
{
// Adding a new pile
pile++;
int size = 0;
HashSet toRemove = new HashSet<>();
// Traverse all piles in increasing
// order of capacities
for (HashMap.Entry tm : occ.entrySet())
{
int mx = tm.getKey();
int ct = tm.getValue();
// Number of boxes of capacity mx
// that can be added to current pile
int use = Math.min(ct, mx - size + 1);
// Update the occurrence
occ.put(mx, occ.get(mx) - use);
// Update the size of the pile
size += use;
if (occ.get(mx) == 0)
toRemove.add(mx);
}
// Remove capacities that are
// no longer available
for (int tm : toRemove)
occ.remove(tm);
}
return pile;
}
// Driver Code
public static void main(String[] args)
{
int[] a = { 0, 0, 1, 1, 2 };
int n = a.length;
System.out.println(countPiles(n, a));
}
}
// This code is contributed by
// sanjeev2552 Python3
# Python3 implementation of the approach
# Function to return the count
# of minimum stacks
def countPiles(n, a):
# Keep track of occurrence
# of each capacity
occ = dict()
# Fill the occurrence map
for i in a:
if i in occ.keys():
occ[i] += 1
else:
occ[i] = 1
# Number of piles is 0 initially
pile = 0
# Traverse occurrences in increasing
# order of capacities.
while (len(occ) > 0):
# Adding a new pile
pile += 1
size = 0
toRemove = dict()
# Traverse all piles in increasing
# order of capacities
for tm in occ:
mx = tm
ct = occ[tm]
# Number of boxes of capacity mx
# that can be added to current pile
use = min(ct, mx - size + 1)
# Update the occurrence
occ[mx] -= use
# Update the size of the pile
size += use
if (occ[mx] == 0):
toRemove[mx] = 1
# Remove capacities that are
# no longer available
for tm in toRemove:
del occ[tm]
return pile
# Driver code
a = [0, 0, 1, 1, 2]
n = len(a)
print(countPiles(n, a))
# This code is contributed
# by Mohit KumarC#
// C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
class GFG
{
// Function to return the count
// of minimum stacks
static int countPiles(int n, int[] a)
{
// Keep track of occurrence
// of each capacity
Dictionary occ = new Dictionary();
// Fill the occurrence map
for (int i = 0; i < n; i++)
{
if(!occ.ContainsKey(a[i]))
{
occ[a[i]]=0;
}
occ[a[i]]++;
}
// Number of piles is 0 initially
int pile = 0;
// Traverse occurrences in increasing
// order of capacities.
while(occ.Count!=0)
{
// Adding a new pile
pile++;
int size = 0;
HashSet toRemove = new HashSet();
Dictionary tmp = occ;
// Traverse all piles in increasing
// order of capacities
foreach(var tm in occ.Keys.ToList())
{
int mx = tm;
int ct = occ[tm];
// Number of boxes of capacity mx
// that can be added to current pile
int use = Math.Min(ct, mx - size + 1);
// Update the occurrence
occ[mx]-= use;
// Update the size of the pile
size += use;
if (occ[mx] == 0)
toRemove.Add(mx);
}
occ = tmp;
// Remove capacities that are
// no longer available
foreach(int tm in toRemove.ToList())
occ.Remove(tm);
}
return pile;
}
// Driver Code
public static void Main(string[] args)
{
int[] a = { 0, 0, 1, 1, 2 };
int n = a.Length;
Console.WriteLine(countPiles(n, a));
}
}
//// This code is contributed by rutvik_56. 输出：
2时间复杂度： O(NlogN)