检查数组中的每对 1 是否至少相距 K 长度

给定一个二进制数组和一个整数 K ，检查数组中的每对 1 是否至少相距 K 长度。如果条件成立则返回真，否则返回假。
例子：

Input: arr = [1, 0, 0, 0, 1, 0, 0, 1, 0, 0], K = 2. 
Output: True 
Explanation: 
Every 1 in the array is at least K distance apart from each other.
Input: [1, 0, 1, 0, 1, 1], K = 1 
Output: False 
Explanation: 
The fifth 1 and sixth 1 are not apart from each other. Hence, the output is false. 
 

方法：
为了解决上面提到的问题，我们必须检查每对相邻 1 之间的距离。找到 1 的第一个位置，然后遍历数组的其余部分，如果为 0，则增加距离，否则，如果距离小于 k，则执行检查操作并将计数再次重置为 0。
下面是上述方法的实现：

C++

// C++ implementation to Check if every pair of 1 in
// the array is at least K length from each other
 
#include 
using namespace std;
 
// Function to check distance
bool kLengthApart(vector& nums, int k)
{
    // Find first position of 1
    int pos = 0, count = 0;
 
    while (pos < nums.size() && nums[pos] == 0)
        pos++;
 
    // Iterate through the rest of array
    for (int i = pos + 1; i < nums.size(); i++) {
        // Increment distance if its 0
        if (nums[i] == 0)
            count++;
 
        // Check if the distance is less than k
        else {
            if (count < k)
                return false;
 
            // Reset count to 0
            count = 0;
        }
    }
 
    // Return the result
    return true;
}
 
// Driver code
int main()
{
    vector nums = { 1, 0, 0, 0, 1, 0, 0, 1, 0, 0 };
    int k = 2;
 
    bool ans = kLengthApart(nums, k);
    if (ans == 1)
        cout << "True" << endl;
 
    else
        cout << "False" << endl;
 
    return 0;
}

Java

// Java implementation to check if
// every pair of 1 in the array is
// at least K length from each other
class Main{
     
// Function to check distance
public static boolean kLengthApart(int[] nums,
                                   int k)
{
     
    // Find first position of 1
    int pos = 0, count = 0;
 
    while (pos < nums.length && nums[pos] == 0)
        pos++;
 
    // Iterate through the rest of array
    for(int i = pos + 1; i < nums.length; i++)
    {
         
       // Increment distance if its 0
       if (nums[i] == 0)
           count++;
            
       // Check if the distance is less than k
       else
       {
           if (count < k)
               return false;
                
           // Reset count to 0
           count = 0;
       }
    }
     
    // Return the result
    return true;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] nums = { 1, 0, 0, 0, 1,
                   0, 0, 1, 0, 0 };
    int k = 2;
    boolean ans = kLengthApart(nums, k);
     
    if (ans)
        System.out.println("True");
 
    else
        System.out.println("False");
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation to check if
# every pair of 1 in the array is
# at least K length from each other
 
# Function to check distance
def kLengthApart(nums, k):
     
    # Find first position of 1
    pos = 0
    count = 0
     
    while (pos < len(nums) and nums[pos] == 0):
        pos += 1
         
    # Iterate through the rest of list
    for i in range(pos + 1, len(nums)):
         
        # Increment distance if its 0
        if nums[i] == 0:
            count += 1
             
        # Check if the distance is less than k
        else :
            if count < k:
                return False
                 
            # Reset count to 0
            count = 0
             
        # Return the result
    return True
     
# Driver Code
if __name__ == "__main__":
      
    nums = [ 1, 0, 0, 0, 1, 0, 0, 1, 0, 0 ]
    k = 2
     
    print(kLengthApart(nums, k))
 
# This code is contributed by rutvik_56

C#

// C# implementation to check if
// every pair of 1 in the array is
// at least K length from each other
using System;
 
class GFG{
     
// Function to check distance
public static bool kLengthApart(int[] nums,
                                int k)
{
     
    // Find first position of 1
    int pos = 0, count = 0;
 
    while (pos < nums.Length && nums[pos] == 0)
        pos++;
 
    // Iterate through the rest of array
    for(int i = pos + 1; i < nums.Length; i++)
    {
        
       // Increment distance if its 0
       if (nums[i] == 0)
           count++;
            
       // Check if the distance is
       // less than k
       else
       {
           if (count < k)
               return false;
            
           // Reset count to 0
           count = 0;
       }
    }
     
    // Return the result
    return true;
}
 
// Driver Code
public static void Main()
{
    int[] nums = { 1, 0, 0, 0, 1,
                   0, 0, 1, 0, 0 };
    int k = 2;
    bool ans = kLengthApart(nums, k);
     
    if (ans)
        Console.Write("True");
    else
        Console.Write("False");
}
}
 
// This code is contributed by chitranayal

Javascript

 
// JavaScript implementation to
// Check if every pair of 1 in
// the array is at least K length
// from each other
 
// Function to check distance
function kLengthApart(nums, k)
{
    // Find first position of 1
    var pos = 0, count = 0;
    var i;
    while (pos < nums.length && nums[pos] == 0)
        pos++;
 
    // Iterate through the rest of array
    for (i = pos + 1; i < nums.length; i++) {
        // Increment distance if its 0
        if (nums[i] == 0)
            count++;
 
        // Check if the distance is less than k
        else {
            if (count < k)
                return false;
 
            // Reset count to 0
            count = 0;
        }
    }
 
    // Return the result
    return true;
}
 
// Driver code
    var nums = [1, 0, 0, 0, 1, 0, 0, 1, 0, 0];
    var k = 2;
 
    var ans = kLengthApart(nums, k);
    if (ans == 1)
        document.write("True");
 
    else
        document.write("False");
 

输出：

True

时间复杂度： O(n)
辅助空间： O(1)

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