📜  最大乘积子数组的长度

📅  最后修改于: 2021-05-14 01:40:06             🧑  作者: Mango

给定大小为N的整数数组arr [] ,任务是查找元素元素乘积为非零的最大长度子数组。 。
例子:

方法:

  1. 保存输入数组中所有零的索引。
  2. 最长的子数组必须位于以下三个范围内:
    • 从零索引开始,到第一个零索引– 1。
    • 介于两个零索引之间。
    • 从最后一个零索引+ 1开始,到N-1结束。
  3. 最终找到所有情况下的最大长度。

这是上述方法的实现:

C++
// C++ program to find maximum 
// length subarray having non
// zero product
#include
using namespace std;
  
// Function that returns the
// maximum length subarray 
// having non zero product
void Maxlength(int arr[],int N)
{
    vector zeroindex;
    int maxlen;
      
    // zeroindex list to store indexex 
    // of zero
    for(int i = 0; i < N; i++)
    {
        if(arr[i] == 0)
            zeroindex.push_back(i);
    }
              
    if(zeroindex.size() == 0)
    {
          
        // If zeroindex list is empty
        // then Maxlength is as 
        // size of array
        maxlen = N;
    }
      
    // If zeroindex list is not empty 
    else
    {
          
        // for example list 1 1 0 0 1
        // is on indexex 0 1 2 3 4
          
        // first zero is on index 2
        // that means two numbers positive,
        // before index 2 so as 
        // their product is positive to
        maxlen = zeroindex[0];
          
        // Checking for other indexex 
        for(int i = 0; 
                i < zeroindex.size() - 1; i++)
        {
              
            // If the difference is greater
            // than maxlen then maxlen 
            // is updated
            if(zeroindex[i + 1]- 
               zeroindex[i] - 1 > maxlen)
            {
                maxlen = zeroindex[i + 1] - 
                         zeroindex[i] - 1;
            }
        }
          
        // To check the length of remaining
        // array after last zeroindex
        if(N - zeroindex[zeroindex.size() - 1] -
           1 > maxlen)
        {
            maxlen = N - zeroindex[
                         zeroindex.size() - 1] - 1;
        }
    }
    cout << maxlen << endl;
}
  
// Driver Code
int main()
{
    int N = 9;
    int arr[] = {7, 1, 0, 1, 2, 0, 9, 2, 1};
      
    Maxlength(arr, N);
}
      
// This code is contributed by Surendra_Gangwar


Java
// Java program to find maximum 
// length subarray having non 
// zero product 
import java.util.*; 
  
class GFG{
  
// Function that returns the 
// maximum length subarray 
// having non zero product 
static void Maxlength(int arr[],int N) 
{ 
    Vector zeroindex = new Vector(); 
      
    int maxlen; 
      
    // zeroindex list to store indexex 
    // of zero 
    for(int i = 0; i < N; i++) 
    { 
        if (arr[i] == 0) 
            zeroindex.add (i); 
    } 
              
    if (zeroindex.size() == 0) 
    { 
          
        // If zeroindex list is empty 
        // then Maxlength is as 
        // size of array 
        maxlen = N; 
    } 
      
    // If zeroindex list is not empty 
    else
    { 
          
        // for example list 1 1 0 0 1 
        // is on indexex 0 1 2 3 4 
          
        // first zero is on index 2 
        // that means two numbers positive, 
        // before index 2 so as 
        // their product is positive to 
        maxlen = (int)zeroindex.get(0); 
          
        // Checking for other indexex 
        for(int i = 0; 
                i < zeroindex.size() - 1; i++) 
        { 
              
            // If the difference is greater 
            // than maxlen then maxlen 
            // is updated 
            if ((int)zeroindex.get(i + 1) - 
                (int)zeroindex.get(i) - 1 > maxlen) 
            { 
                maxlen = (int)zeroindex.get(i + 1) - 
                         (int)zeroindex.get(i) - 1; 
            } 
        } 
          
        // To check the length of remaining 
        // array after last zeroindex 
        if (N - (int)zeroindex.get(
                     zeroindex.size() - 1) - 
                                    1 > maxlen) 
        { 
            maxlen = N - (int)zeroindex.get( 
                              zeroindex.size() - 1) - 1; 
        } 
    } 
    System.out.println(maxlen); 
} 
  
// Driver code 
public static void main(String args[]) 
{ 
    int N = 9; 
    int arr[] = { 7, 1, 0, 1, 2, 0, 9, 2, 1 }; 
      
    Maxlength(arr, N); 
}
}
  
// This code is contributed by amreshkumar3


Python3
# Python3 program to find 
# maximum length subarray 
# having non zero product
  
# function that returns the
# maximum length subarray 
# having non zero product
def Maxlength(arr, N):
      
    zeroindex =[]
      
    # zeroindex list to store indexex 
    # of zero
    for i in range(N):
        if(arr[i] == 0):
            zeroindex.append(i)
              
      
    if(len(zeroindex) == 0):
        # if zeroindex list is empty
        # then Maxlength is as 
        # size of array
        maxlen = N
    # if zeroindex list is not empty 
    else:
        # for example list 1 1 0 0 1
        # is on indexex 0 1 2 3 4
          
        # first zero is on index 2
        # that means two numbers positive,
        # before index 2 so as 
        # their product is positive to
          
        maxlen = zeroindex[0]
          
        # checking for other indexex 
        for i in range(0, len(zeroindex)-1):
              
            # if the difference is greater
            # than maxlen then maxlen 
            # is updated
            if(zeroindex[i + 1]\
               - zeroindex[i] - 1\
               > maxlen):
                maxlen = zeroindex[i + 1]\
                         - zeroindex[i] - 1
              
          
        # to check the length of remaining
        # array after last zeroindex
        if(N - zeroindex[len(zeroindex) - 1]\
                                 - 1 > maxlen):
            maxlen = N\
             - zeroindex[len(zeroindex) - 1] - 1
      
    print(maxlen)
  
  
# Driver Code
if __name__ == "__main__":
    N = 9
    arr = [7, 1, 0, 1, 2, 0, 9, 2, 1]
    Maxlength(arr, N)


C#
// C# program to find maximum 
// length subarray having non 
// zero product 
using System; 
using System.Collections.Generic;
  
class GFG{
  
// Function that returns the 
// maximum length subarray 
// having non zero product 
static void Maxlength(int []arr,int N) 
{ 
    int[] zeroindex = new int[20000]; 
    int maxlen; 
      
    // zeroindex list to store indexex 
    // of zero 
    int size = 0;
    for(int i = 0; i < N; i++) 
    { 
        if (arr[i] == 0) 
            zeroindex[size++] = i; 
    } 
              
    if (size == 0) 
    { 
          
        // If zeroindex list is empty 
        // then Maxlength is as 
        // size of array 
        maxlen = N; 
    } 
      
    // If zeroindex list is not empty 
    else
    { 
          
        // for example list 1 1 0 0 1 
        // is on indexex 0 1 2 3 4 
          
        // first zero is on index 2 
        // that means two numbers positive, 
        // before index 2 so as 
        // their product is positive to 
        maxlen = zeroindex[0]; 
          
        // Checking for other indexex
        for(int i = 0; i < size; i++) 
        { 
              
            // If the difference is greater 
            // than maxlen then maxlen 
            // is updated 
            if (zeroindex[i + 1]- 
                zeroindex[i] - 1 > maxlen)
            { 
                maxlen = zeroindex[i + 1] - 
                         zeroindex[i] - 1; 
            } 
        } 
          
        // To check the length of remaining 
        // array after last zeroindex 
        if (N - zeroindex[size - 1] - 1 > maxlen) 
        { 
            maxlen = N - zeroindex[size - 1] - 1; 
        } 
    } 
    Console.WriteLine(maxlen); 
} 
  
// Driver code 
public static void Main() 
{ 
    int N = 9; 
    int []arr = { 7, 1, 0, 1, 2, 0, 9, 2, 1 }; 
      
    Maxlength(arr, N); 
}
}
  
// This code is contributed by amreshkumar3


输出:
3

时间复杂度: O(N)
辅助空间: O(N)