零之间数组中所有元素的总和

给定一个由N 个整数组成的数组arr[] ，任务是找到给定数组中两个零之间所有元素的总和。如果可能，则打印所有总和，否则打印“-1” 。
注意：给定数组中没有连续的零。

例子：

Input: arr[] = { 1, 0, 3, 4, 0, 4, 4, 0, 2, 1, 4, 0, 3 } 
Output: 7 8 7 
Explanation: 
The sum of elements between every zero are: 
3 + 4 = 7 
4 + 4 = 8 
2 + 1 + 4 = 7

Input: arr[] = { 1, 3, 4, 6, 0} 
Output: -1  

方法：

1. 遍历给定的数组arr[]并找到元素为 0 的第一个索引。
2. 如果出现任何值为 0 的元素，则开始将其后的元素总和存储在向量中(比如A[] )，直到出现下一个零。
3. 对出现的每个零重复上述步骤。
4. 打印存储在 A[] 中的元素。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include "bits/stdc++.h"
using namespace std;
 
// Function to find the sum between two
// zeros in the given array arr[]
void sumBetweenZero(int arr[], int N)
{
 
    int i = 0;
 
    // To store the sum of element
    // between two zeros
    vector A;
 
    // To store the sum
    int sum = 0;
 
    // Find first index of 0
    for (i = 0; i < N; i++) {
        if (arr[i] == 0) {
            i++;
            break;
        }
    }
 
    // Traverse the given array arr[]
    for (; i < N; i++) {
 
        // If 0 occurs then add it to A[]
        if (arr[i] == 0) {
            A.push_back(sum);
            sum = 0;
        }
 
        // Else add element to the sum
        else {
            sum += arr[i];
        }
    }
 
    // Print all the sum stored in A
    for (int i = 0; i < A.size(); i++) {
        cout << A[i] << ' ';
    }
 
    // If there is no such element print -1
    if (A.size() == 0)
        cout << "-1";
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 0, 3, 4, 0, 4, 4,
                  0, 2, 1, 4, 0, 3 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    sumBetweenZero(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the sum between two
// zeros in the given array arr[]
static void sumBetweenZero(int arr[], int N)
{
    int i = 0;
 
    // To store the sum of element
    // between two zeros
    Vector A = new Vector();
 
    // To store the sum
    int sum = 0;
 
    // Find first index of 0
    for(i = 0; i < N; i++)
    {
       if (arr[i] == 0)
       {
           i++;
           break;
       }
    }
 
    // Traverse the given array arr[]
    for(; i < N; i++)
    {
        
       // If 0 occurs then add it to A[]
       if (arr[i] == 0)
       {
           A.add(sum);
           sum = 0;
       }
        
       // Else add element to the sum
       else
       {
           sum += arr[i];
       }
    }
 
    // Print all the sum stored in A
    for(int j = 0; j < A.size(); j++)
    {
       System.out.print(A.get(j) + " ");
    }
 
    // If there is no such element print -1
    if (A.size() == 0)
        System.out.print("-1");
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 0, 3, 4, 0, 4, 4,
                  0, 2, 1, 4, 0, 3 };
    int N = arr.length;
 
    // Function call
    sumBetweenZero(arr, N);
}
}
 
// This code is contributed by gauravrajput1

Python3

#Python3 program for the above approach
 
# Function to find the sum between two
# zeros in the given array arr[]
def sumBetweenZero(arr, N):
    i = 0
 
    # To store the sum of the element
    # between two zeros
    A = []
 
    # To store the sum
    sum = 0
 
    # Find first index of 0
    for i in range(N):
        if (arr[i] == 0):
            i += 1
            break
    k = i
 
    # Traverse the given array arr[]
    for i in range(k, N, 1):
         
        # If 0 occurs then add it to A[]
        if (arr[i] == 0):
            A.append(sum)
            sum = 0
 
        # Else add element to the sum
        else:
            sum += arr[i]
 
    # Print all the sum stored in A
    for i in range(len(A)):
        print(A[i], end = ' ')
 
    # If there is no such element print -1
    if (len(A) == 0):
        print("-1")
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 0, 3, 4, 0, 4, 4,
            0, 2, 1, 4, 0, 3 ]
 
    N = len(arr)
 
    # Function call
    sumBetweenZero(arr, N)
 
# This code is contributed by Bhupendra_Singh

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the sum between two
// zeros in the given array []arr
static void sumBetweenZero(int []arr, int N)
{
    int i = 0;
 
    // To store the sum of element
    // between two zeros
    List A = new List();
 
    // To store the sum
    int sum = 0;
 
    // Find first index of 0
    for(i = 0; i < N; i++)
    {
       if (arr[i] == 0)
       {
           i++;
           break;
       }
    }
 
    // Traverse the given array []arr
    for(; i < N; i++)
    {
         
       // If 0 occurs then add it to []A
       if (arr[i] == 0)
       {
           A.Add(sum);
           sum = 0;
       }
        
       // Else add element to the sum
       else
       {
           sum += arr[i];
       }
    }
     
    // Print all the sum stored in A
    for(int j = 0; j < A.Count; j++)
    {
       Console.Write(A[j] + " ");
    }
 
    // If there is no such element print -1
    if (A.Count == 0)
        Console.Write("-1");
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 0, 3, 4, 0, 4, 4,
                  0, 2, 1, 4, 0, 3 };
    int N = arr.Length;
 
    // Function call
    sumBetweenZero(arr, N);
}
}
 
// This code is contributed by gauravrajput1

Javascript

// Javascript program for the above approach
 
 
// Function to find the sum between two
// zeros in the given array arr[]
function sumBetweenZero(arr, N) {
 
    let i = 0;
 
    // To store the sum of element
    // between two zeros
    let A = new Array();
 
    // To store the sum
    let sum = 0;
 
    // Find first index of 0
    for (i = 0; i < N; i++) {
        if (arr[i] == 0) {
            i++;
            break;
        }
    }
 
    // Traverse the given array arr[]
    for (; i < N; i++) {
 
        // If 0 occurs then add it to A[]
        if (arr[i] == 0) {
            A.push(sum);
            sum = 0;
        }
 
        // Else add element to the sum
        else {
            sum += arr[i];
        }
    }
 
    // Print all the sum stored in A
    for (let i = 0; i < A.length; i++) {
        document.write(A[i] + ' ');
    }
 
    // If there is no such element print -1
    if (A.length == 0)
        document.write("-1");
}
 
// Driver Code
 
let arr = [1, 0, 3, 4, 0, 4, 4,
    0, 2, 1, 4, 0, 3];
 
let N = arr.length;
 
// Function call
sumBetweenZero(arr, N);
 
// This code is contributed by _saurabh_jaiswal

输出：

7 8 7

时间复杂度： O(N) ，其中 N 是数组的长度。

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