根据给定条件分配给儿童所需的最少糖果数量

给定一个由N 个正整数组成的数组arr[] ，代表N 个孩子的评分，任务是找到分发给N 个孩子所需的最少糖果数量，使得每个孩子至少得到一个糖果，并且评分较高的孩子得到比邻居更多的糖果。

例子：

Input: arr[] = {1, 0, 2}
Output: 5
Explanation:
Consider the distribution of candies as {2, 1, 2} that satisfy the given conditions. Therefore, the sum of candies is 2 + 1 + 2 = 5, which is the minimum required candies.

Input: arr[] = {1, 2, 2}
Output: 4

方法：可以使用贪心方法解决给定的问题。请按照以下步骤解决问题：

• 初始化一个数组，比如ans[]来存储分配给每个孩子的糖果数量，并将其初始化为1到每个数组元素ans[] 。
• 遍历给定的数组arr[]并且如果arr[i + 1]的值大于arr[i]并且ans[i + 1] 的值至少是ans[i] ，则更新ans[的值i + 1]作为ans[i] + 1 。
• 从后面遍历给定数组并执行以下步骤：
  • 如果ARR的值[i]是比ARR大于第[i + 1]和ANS [I]的值是至少ANS第[i + 1]，然后更新ANS的值[I + 1]如ANS [I ] + 1 。
  • 如果ARR [I]的值等于给Arr第[i + 1]和ANS [I]的值小于ANS第[i + 1]，然后更新ANS的值[I + 1]如ANS [I ] + 1 。
• 完成上述步骤后，打印数组ans[]的总和作为结果糖果的总和。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the minimum number
// of candies that is to be distributed
int countCandies(int arr[], int n)
{
    // Stores total count of candies
    int sum = 0;
 
    // Stores the amount of candies
    // allocated to a student
    int ans[n];
 
    // If the value of N is 1
    if (n == 1) {
        return 1;
    }
 
    // Initialize with 1 all array
    // element
    for (int i = 0; i < n; i++)
        ans[i] = 1;
 
    // Traverse the array arr[]
    for (int i = 0; i < n - 1; i++) {
 
        // If arr[i+1] is greater than
        // arr[i] and ans[i+1] is
        // at most ans[i]
        if (arr[i + 1] > arr[i]
            && ans[i + 1] <= ans[i]) {
 
            // Assign ans[i]+1 to
            // ans[i+1]
            ans[i + 1] = ans[i] + 1;
        }
    }
 
    // Iterate until i is atleast 0
    for (int i = n - 2; i >= 0; i--) {
 
        // If arr[i] is greater than
        // arr[i+1] and ans[i] is
        // at most ans[i+1]
        if (arr[i] > arr[i + 1]
            && ans[i] <= ans[i + 1]) {
 
            // Assign ans[i+1]+1 to
            // ans[i]
            ans[i] = ans[i + 1] + 1;
        }
 
        // If arr[i] is equals arr[i+1]
        // and ans[i] is less than
        // ans[i+1]
        else if (arr[i] == arr[i + 1]
                 && ans[i] < ans[i + 1]) {
 
            // Assign ans[i+1]+1 to
            // ans[i]
            ans[i] = ans[i + 1] + 1;
        }
 
        // Increment the sum by ans[i]
        sum += ans[i];
    }
 
    sum += ans[n - 1];
 
    // Return the resultant sum
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 0, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << countCandies(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to count the minimum number
// of candies that is to be distributed
static int countCandies(int arr[], int n)
{
     
    // Stores total count of candies
    int sum = 0;
     
    // Stores the amount of candies
    // allocated to a student
    int[] ans = new int[n];
     
    // If the value of N is 1
    if (n == 1)
    {
        return 1;
    }
     
    // Initialize with 1 all array
    // element
    for(int i = 0; i < n; i++)
        ans[i] = 1;
     
    // Traverse the array arr[]
    for(int i = 0; i < n - 1; i++)
    {
     
        // If arr[i+1] is greater than
        // arr[i] and ans[i+1] is
        // at most ans[i]
        if (arr[i + 1] > arr[i] &&
            ans[i + 1] <= ans[i])
        {
         
            // Assign ans[i]+1 to
            // ans[i+1]
            ans[i + 1] = ans[i] + 1;
        }
    }
     
    // Iterate until i is atleast 0
    for(int i = n - 2; i >= 0; i--)
    {
     
        // If arr[i] is greater than
        // arr[i+1] and ans[i] is
        // at most ans[i+1]
        if (arr[i] > arr[i + 1] &&
            ans[i] <= ans[i + 1])
        {
         
            // Assign ans[i+1]+1 to
            // ans[i]
            ans[i] = ans[i + 1] + 1;
        }
         
        // If arr[i] is equals arr[i+1]
        // and ans[i] is less than
        // ans[i+1]
        else if (arr[i] == arr[i + 1] &&
                 ans[i] < ans[i + 1])
        {
         
            // Assign ans[i+1]+1 to
            // ans[i]
            ans[i] = ans[i + 1] + 1;
        }
         
        // Increment the sum by ans[i]
        sum += ans[i];
    }
     
    sum += ans[n - 1];
     
    // Return the resultant sum
    return sum;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 0, 2 };
    int N = arr.length;
     
    System.out.println(countCandies(arr, N));
}
}
 
// This code is contributed by patel2127

Python3

# Python3 program for the above approach
 
# Function to count the minimum number
# of candies that is to be distributed
def countCandies(arr, n):
     
    # Stores total count of candies
    sum = 0
 
    # Stores the amount of candies
    # allocated to a student
    ans = [1] * n
 
    # If the value of N is 1
    if (n == 1):
        return 1
 
    # Initialize with 1 all array
    # element
    # for (i = 0 i < n i++)
    #     ans[i] = 1
 
    # Traverse the array arr[]
    for i in range(n - 1):
 
        # If arr[i+1] is greater than
        # arr[i] and ans[i+1] is
        # at most ans[i]
        if (arr[i + 1] > arr[i] and
            ans[i + 1] <= ans[i]):
             
            # Assign ans[i]+1 to
            # ans[i+1]
            ans[i + 1] = ans[i] + 1
 
    # Iterate until i is atleast 0
    for i in range(n - 2, -1, -1):
         
        # If arr[i] is greater than
        # arr[i+1] and ans[i] is
        # at most ans[i+1]
        if (arr[i] > arr[i + 1] and
            ans[i] <= ans[i + 1]):
 
            # Assign ans[i+1]+1 to
            # ans[i]
            ans[i] = ans[i + 1] + 1
 
        # If arr[i] is equals arr[i+1]
        # and ans[i] is less than
        # ans[i+1]
        elif (arr[i] == arr[i + 1] and
              ans[i] < ans[i + 1]):
 
            # Assign ans[i+1]+1 to
            # ans[i]
            ans[i] = ans[i + 1] + 1
 
        # Increment the sum by ans[i]
        sum += ans[i]
 
    sum += ans[n - 1]
 
    # Return the resultant sum
    return sum
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 0, 2 ]
    N = len(arr)
     
    print (countCandies(arr, N))
 
# This code is contributed by mohit kumar 29

C#

using System;
 
public class GFG{
     
    // Function to count the minimum number
// of candies that is to be distributed
static int countCandies(int[] arr, int n)
{
      
    // Stores total count of candies
    int sum = 0;
      
    // Stores the amount of candies
    // allocated to a student
    int[] ans = new int[n];
      
    // If the value of N is 1
    if (n == 1)
    {
        return 1;
    }
      
    // Initialize with 1 all array
    // element
    for(int i = 0; i < n; i++)
        ans[i] = 1;
      
    // Traverse the array arr[]
    for(int i = 0; i < n - 1; i++)
    {
      
        // If arr[i+1] is greater than
        // arr[i] and ans[i+1] is
        // at most ans[i]
        if (arr[i + 1] > arr[i] &&
            ans[i + 1] <= ans[i])
        {
          
            // Assign ans[i]+1 to
            // ans[i+1]
            ans[i + 1] = ans[i] + 1;
        }
    }
      
    // Iterate until i is atleast 0
    for(int i = n - 2; i >= 0; i--)
    {
      
        // If arr[i] is greater than
        // arr[i+1] and ans[i] is
        // at most ans[i+1]
        if (arr[i] > arr[i + 1] &&
            ans[i] <= ans[i + 1])
        {
          
            // Assign ans[i+1]+1 to
            // ans[i]
            ans[i] = ans[i + 1] + 1;
        }
          
        // If arr[i] is equals arr[i+1]
        // and ans[i] is less than
        // ans[i+1]
        else if (arr[i] == arr[i + 1] &&
                 ans[i] < ans[i + 1])
        {
          
            // Assign ans[i+1]+1 to
            // ans[i]
            ans[i] = ans[i + 1] + 1;
        }
          
        // Increment the sum by ans[i]
        sum += ans[i];
    }
      
    sum += ans[n - 1];
      
    // Return the resultant sum
    return sum;
}
  
// Driver Code
     
    static public void Main (){
         
        int[] arr = { 1, 0, 2 };
    int N = arr.Length;
      
    Console.WriteLine(countCandies(arr, N));
         
    }
}
 
// This code is contributed by rag2127

Javascript

 
// Javascript program for the above approach
 
// Function to count the minimum number
// of candies that is to be distributed
function countCandies(arr, n)
{
     
    // Stores total count of candies
    let sum = 0;
  
    // Stores the amount of candies
    // allocated to a student
    let ans = new Array(n);
  
    // If the value of N is 1
    if (n == 1)
    {
        return 1;
    }
  
    // Initialize with 1 all array
    // element
    for(let i = 0; i < n; i++)
        ans[i] = 1;
  
    // Traverse the array arr[]
    for(let i = 0; i < n - 1; i++)
    {
         
        // If arr[i+1] is greater than
        // arr[i] and ans[i+1] is
        // at most ans[i]
        if (arr[i + 1] > arr[i] &&
            ans[i + 1] <= ans[i])
        {
             
            // Assign ans[i]+1 to
            // ans[i+1]
            ans[i + 1] = ans[i] + 1;
        }
    }
  
    // Iterate until i is atleast 0
    for(let i = n - 2; i >= 0; i--)
    {
         
        // If arr[i] is greater than
        // arr[i+1] and ans[i] is
        // at most ans[i+1]
        if (arr[i] > arr[i + 1] &&
            ans[i] <= ans[i + 1])
        {
             
            // Assign ans[i+1]+1 to
            // ans[i]
            ans[i] = ans[i + 1] + 1;
        }
  
        // If arr[i] is equals arr[i+1]
        // and ans[i] is less than
        // ans[i+1]
        else if (arr[i] == arr[i + 1] &&
                 ans[i] < ans[i + 1])
        {
             
            // Assign ans[i+1]+1 to
            // ans[i]
            ans[i] = ans[i + 1] + 1;
        }
  
        // Increment the sum by ans[i]
        sum += ans[i];
    }
    sum += ans[n - 1];
  
    // Return the resultant sum
    return sum;
}
 
// Driver Code
let arr = [ 1, 0, 2 ];
let N = arr.length;
 
document.write(countCandies(arr, N))
 
// This code is contributed by unknown2108
 

输出：

5

时间复杂度： O(N)
辅助空间： O(N)

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