以矩阵的螺旋形式打印第 K 个元素

给定一个 n X m 阶的二维矩阵，以矩阵的螺旋形式打印第 K 个元素。请参阅以下示例。
例子：

Input: mat[][] = 
          {{1, 2, 3, 4}
           {5, 6, 7, 8}
           {9, 10, 11, 12}
           {13, 14, 15, 16}}
       k = 6
Output: 12
Explanation: The elements in spiral order is 
1, 2, 3, 4, 8, 12, 16, 15...
so the 6th element is 12

Input: mat[][] =
       {{1, 2, 3, 4, 5, 6}
        {7, 8, 9, 10, 11, 12}
        {13, 14, 15, 16, 17, 18}}
       k = 17
Output: 10
Explanation: The elements in spiral order is 
1, 2, 3, 4, 5, 6, 12, 18, 17,
16, 15, 14, 13, 7, 8, 9, 10, 11 
so the 17 th element is 10.  

简单方法：一种简单的解决方案是以螺旋形式开始遍历矩阵 Print Spiral Matrix 并启动一个计数器，即； count = 0。每当 count 等于 K 时，打印该元素。

• 算法：
  1. 保持一个变量count = 0来存储计数。
  2. 从头到尾以螺旋方式遍历矩阵。
  3. 每次迭代将计数增加 1。
  4. 如果计数等于 k 的给定值，则打印元素并中断。

执行

C++

#include 
using namespace std;
#define R 3
#define C 6
 
void spiralPrint(int m, int n, int a[R][C], int c)
{
    int i, k = 0, l = 0;
    int count = 0;
 
    /* k - starting row index
        m - ending row index
        l - starting column index
        n - ending column index
        i - iterator
    */
 
    while (k < m && l < n) {
        /* check the first row from
            the remaining rows */
        for (i = l; i < n; ++i) {
            count++;
 
            if (count == c)
                cout << a[k][i] << " ";
        }
        k++;
 
        /* check the last column
        from the remaining columns */
        for (i = k; i < m; ++i) {
            count++;
 
            if (count == c)
                cout << a[i][n - 1] << " ";
        }
        n--;
 
        /* check the last row from
                the remaining rows */
        if (k < m) {
            for (i = n - 1; i >= l; --i) {
                count++;
 
                if (count == c)
                    cout << a[m - 1][i] << " ";
            }
            m--;
        }
 
        /* check the first column from
                the remaining columns */
        if (l < n) {
            for (i = m - 1; i >= k; --i) {
                count++;
 
                if (count == c)
                    cout << a[i][l] << " ";
            }
            l++;
        }
    }
}
 
/* Driver program to test above functions */
int main()
{
    int a[R][C] = { { 1, 2, 3, 4, 5, 6 },
                    { 7, 8, 9, 10, 11, 12 },
                    { 13, 14, 15, 16, 17, 18 } },
        k = 17;
 
    spiralPrint(R, C, a, k);
    return 0;
}

Java

import java.io.*;
 
class GFG {
    static int R = 3;
    static int C = 6;
 
    static void spiralPrint(int m, int n, int[][] a, int c)
    {
        int i, k = 0, l = 0;
        int count = 0;
     
        /* k - starting row index
            m - ending row index
            l - starting column index
            n - ending column index
            i - iterator
        */
     
        while (k < m && l < n) {
            /* check the first row from
                the remaining rows */
            for (i = l; i < n; ++i) {
                count++;
     
                if (count == c)
                    System.out.println(a[k][i]+" ");
            }
            k++;
     
            /* check the last column
            from the remaining columns */
            for (i = k; i < m; ++i) {
                count++;
     
                if (count == c)
                    System.out.println(a[i][n - 1]+" ");
            }
            n--;
     
            /* check the last row from
                    the remaining rows */
            if (k < m) {
                for (i = n - 1; i >= l; --i) {
                    count++;
     
                    if (count == c)
                         
                    System.out.println(a[m - 1][i]+" ");
                }
                m--;
            }
     
            /* check the first column from
                    the remaining columns */
            if (l < n) {
                for (i = m - 1; i >= k; --i) {
                    count++;
     
                    if (count == c)
                        System.out.println(a[i][l]+" ");
                }
                l++;
            }
        }
    }
     
    /* Driver program to test above functions */
    public static void main (String[] args)
    {
        int a[][] = { { 1, 2, 3, 4, 5, 6 },
                        { 7, 8, 9, 10, 11, 12 },
                        { 13, 14, 15, 16, 17, 18 } };
        int k = 17;
     
        spiralPrint(R, C, a, k);
    }
}
 
// This code is contributed by shivanisinghss2110

Python3

R = 3
C = 6
 
def spiralPrint(m, n, a, c):
    k = 0
    l = 0
    count = 0
    """ k - starting row index
    m - ending row index
    l - starting column index
    n - ending column index
    i - iterator
    """
    while (k < m and l < n):
        for i in range(l,n):
            count+=1
             
            if (count == c):
                print(a[k][i] , end=" ")
         
        k+=1
        """ check the last column
        from the remaining columns """
        for i in range(k,m):
            count+=1
            if (count == c):
                print(a[i][n - 1],end=" ")
        n-=1
        """ check the last row from
        the remaining rows """
        if (k < m):
            for i in range(n - 1,l-1,-1):
                count+=1
                if (count == c):
                    print(a[m - 1][i],end=" ")
            m-=1
        """ check the first column from
        the remaining columns """
        if (l < n):
            for i in range(m - 1,k-1,-1):
                count+=1
                if (count == c):
                    print(a[i][l],end=" ")
            l+=1
 
""" Driver program to test above functions """
 
a = [[1, 2, 3, 4, 5, 6 ],[ 7, 8, 9, 10, 11, 12 ],[ 13, 14, 15, 16, 17, 18]]
k = 17
spiralPrint(R, C, a, k)
 
# This code is contributed by shivanisingh

C#

using System;
class GFG
{
 
  static int R = 3;
  static int C = 6;
  static void spiralPrint(int m, int n,
                          int[,] a, int c) 
  { 
    int i, k = 0, l = 0; 
    int count = 0; 
 
    /* k - starting row index 
            m - ending row index 
            l - starting column index 
            n - ending column index 
            i - iterator 
        */
 
    while (k < m && l < n)
    {
 
      /* check the first row from 
                the remaining rows */
      for (i = l; i < n; ++i)
      { 
        count++; 
 
        if (count == c) 
          Console.WriteLine(a[k, i] + " ");
      } 
      k++; 
 
      /* check the last column 
            from the remaining columns */
      for (i = k; i < m; ++i)
      { 
        count++; 
 
        if (count == c) 
          Console.WriteLine(a[i, n - 1] + " ");
      } 
      n--; 
 
      /* check the last row from 
                    the remaining rows */
      if (k < m)
      { 
        for (i = n - 1; i >= l; --i)
        { 
          count++; 
 
          if (count == c) 
 
            Console.WriteLine(a[m - 1, i] + " ");
        } 
        m--; 
      } 
 
      /* check the first column from 
                    the remaining columns */
      if (l < n)
      { 
        for (i = m - 1; i >= k; --i)
        { 
          count++; 
 
          if (count == c) 
            Console.WriteLine(a[i, l] + " "); 
        } 
        l++; 
      } 
    } 
  }
 
  // Driver code
  static void Main()
  {
    int[,] a = { { 1, 2, 3, 4, 5, 6 }, 
                { 7, 8, 9, 10, 11, 12 }, 
                { 13, 14, 15, 16, 17, 18 } }; 
    int k = 17; 
    spiralPrint(R, C, a, k); 
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

 
    let R = 3;
    let C = 6;
     
    function spiralPrint(m, n, a, c)
    {
        let i, k = 0, l = 0;
        let count = 0;
 
        /* k - starting row index
            m - ending row index
            l - starting column index
            n - ending column index
            i - iterator
        */
 
        while (k < m && l < n) {
            /* check the first row from
                the remaining rows */
            for (i = l; i < n; ++i) {
                count++;
 
                if (count == c)
                    document.write(a[k][i] + " ");
            }
            k++;
 
            /* check the last column
            from the remaining columns */
            for (i = k; i < m; ++i) {
                count++;
 
                if (count == c)
                    document.write(a[i][n - 1] + " ");
            }
            n--;
 
            /* check the last row from
                    the remaining rows */
            if (k < m) {
                for (i = n - 1; i >= l; --i) {
                    count++;
 
                    if (count == c)
                        document.write(a[m - 1][i] + " ");
                }
                m--;
            }
 
            /* check the first column from
                    the remaining columns */
            if (l < n) {
                for (i = m - 1; i >= k; --i) {
                    count++;
 
                    if (count == c)
                        document.write(a[i][l] + " ");
                }
                l++;
            }
        }
    }
   
      let a = [ [ 1, 2, 3, 4, 5, 6 ],
             [ 7, 8, 9, 10, 11, 12 ],
             [ 13, 14, 15, 16, 17, 18 ] ],
        k = 17;
  
    spiralPrint(R, C, a, k);
     

输出：

10

复杂性分析：

• 时间复杂度： O(R*C)，需要单次遍历矩阵，所以时间复杂度为 O(R*C)。
• 空间复杂度： O(1)，需要恒定空间。

有效方法：在以螺旋顺序遍历阵列时，使用循环遍历边。因此，如果可以找到第 k 个元素在给定的一侧，那么可以在恒定时间内找到第 k 个元素。这可以递归地以及迭代地完成。

• 算法 ：
  1. 以螺旋或循环的形式遍历矩阵。
  2. 所以一个循环可以分为 4 个部分，所以如果循环的大小是 m X n。
  3. 元素在第一行，即 k <= m
  4. 元素在最后一列，即 k <= (m+n-1)
  5. 元素在最后一行，即 k <= (m+n-1+m-1)
  6. 元素在第一列，即 k <= (m+n-1+m-1+n-2)
  7. 如果满足上述任何一个条件，则可以找到第 k 个元素是常数时间。
  8. 否则从数组中删除循环并递归调用函数。
• 执行：
  

C++

// C++ program for Kth element in spiral
// form of matrix
#include 
#define MAX 100
using namespace std;
 
/* function for Kth element */
int findK(int A[MAX][MAX], int n, int m, int k)
{
    if (n < 1 || m < 1)
        return -1;
 
    /*....If element is in outermost ring ....*/
    /* Element is in first row */
    if (k <= m)
        return A[0][k - 1];
 
    /* Element is in last column */
    if (k <= (m + n - 1))
        return A[(k - m)][m - 1];
 
    /* Element is in last row */
    if (k <= (m + n - 1 + m - 1))
        return A[n - 1][m - 1 - (k - (m + n - 1))];
 
    /* Element is in first column */
    if (k <= (m + n - 1 + m - 1 + n - 2))
        return A[n - 1 - (k - (m + n - 1 + m - 1))][0];
 
    /*....If element is NOT in outermost ring ....*/
    /* Recursion for sub-matrix. &A[1][1] is
    address to next inside sub matrix.*/
    return findK((int(*)[MAX])(&(A[1][1])), n - 2,
                 m - 2, k - (2 * n + 2 * m - 4));
}
 
/* Driver code */
int main()
{
    int a[MAX][MAX] = { { 1, 2, 3, 4, 5, 6 },
                        { 7, 8, 9, 10, 11, 12 },
                        { 13, 14, 15, 16, 17, 18 } };
    int k = 17;
    cout << findK(a, 3, 6, k) << endl;
    return 0;
}

Java

// Java program for Kth element in spiral
// form of matrix
class GFG {
 
    static int MAX = 100;
 
    /* function for Kth element */
    static int findK(int A[][], int i, int j,
                     int n, int m, int k)
    {
        if (n < 1 || m < 1)
            return -1;
 
        /*.....If element is in outermost ring ....*/
        /* Element is in first row */
        if (k <= m)
            return A[i + 0][j + k - 1];
 
        /* Element is in last column */
        if (k <= (m + n - 1))
            return A[i + (k - m)][j + m - 1];
 
        /* Element is in last row */
        if (k <= (m + n - 1 + m - 1))
            return A[i + n - 1][j + m - 1 - (k - (m + n - 1))];
 
        /* Element is in first column */
        if (k <= (m + n - 1 + m - 1 + n - 2))
            return A[i + n - 1 - (k - (m + n - 1 + m - 1))][j + 0];
 
        /*.....If element is NOT in outermost ring ....*/
        /* Recursion for sub-matrix. &A[1][1] is
    address to next inside sub matrix.*/
        return findK(A, i + 1, j + 1, n - 2,
                     m - 2, k - (2 * n + 2 * m - 4));
    }
 
    /* Driver code */
    public static void main(String args[])
    {
        int a[][] = { { 1, 2, 3, 4, 5, 6 },
                      { 7, 8, 9, 10, 11, 12 },
                      { 13, 14, 15, 16, 17, 18 } };
        int k = 17;
        System.out.println(findK(a, 0, 0, 3, 6, k));
    }
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 program for Kth element in spiral
# form of matrix
MAX = 100
 
''' function for Kth element '''
def findK(A, n, m, k):
     
    if (n < 1 or m < 1):
        return -1
         
    '''....If element is in outermost ring ....'''
    ''' Element is in first row '''
    if (k <= m):
        return A[0][k - 1]
         
    ''' Element is in last column '''
    if (k <= (m + n - 1)):
        return A[(k - m)][m - 1]
         
    ''' Element is in last row '''
    if (k <= (m + n - 1 + m - 1)):
        return A[n - 1][m - 1 - (k - (m + n - 1))]
     
    ''' Element is in first column '''
    if (k <= (m + n - 1 + m - 1 + n - 2)):
        return A[n - 1 - (k - (m + n - 1 + m - 1))][0]
         
         
    '''....If element is NOT in outermost ring ....'''
    ''' Recursion for sub-matrix. &A[1][1] is
    address to next inside sub matrix.'''
    A.pop(0)
    [j.pop(0) for j in A]
    return findK(A, n - 2, m - 2, k - (2 * n + 2 * m - 4))
 
''' Driver code '''
 
a = [[1, 2, 3, 4, 5, 6],[7, 8, 9, 10, 11, 12 ],
    [ 13, 14, 15, 16, 17, 18 ]]
k = 17
print(findK(a, 3, 6, k))
 
# This code is contributed by shivanisinghss2110

C＃

// C# program for Kth element in spiral
// form of matrix
using System;
class GFG
{
 
  /* function for Kth element */
  static int findK(int[,] A, int i, int j,
                   int n, int m, int k)
  {
    if (n < 1 || m < 1)
      return -1;
 
    /*.....If element is in outermost ring ....*/
    /* Element is in first row */
    if (k <= m)
      return A[i + 0, j + k - 1];
 
    /* Element is in last column */
    if (k <= (m + n - 1))
      return A[i + (k - m), j + m - 1];
 
    /* Element is in last row */
    if (k <= (m + n - 1 + m - 1))
      return A[i + n - 1, j + m - 1 - (k - (m + n - 1))];
 
    /* Element is in first column */
    if (k <= (m + n - 1 + m - 1 + n - 2))
      return A[i + n - 1 - (k - (m + n - 1 + m - 1)), j + 0];
 
    /*.....If element is NOT in outermost ring ....*/
    /* Recursion for sub-matrix. &A[1][1] is
    address to next inside sub matrix.*/
    return findK(A, i + 1, j + 1, n - 2,
                 m - 2, k - (2 * n + 2 * m - 4));
  }
 
  // Driver code
  static void Main()
  {
    int[,] a = { { 1, 2, 3, 4, 5, 6 },
                { 7, 8, 9, 10, 11, 12 },
                { 13, 14, 15, 16, 17, 18 } };
    int k = 17;
    Console.WriteLine(findK(a, 0, 0, 3, 6, k));
  }
}
 
// This code is contributed by divyesh072019.

Javascript

 
// JavaScript program for Kth element in spiral
// form of matrix
let MAX = 100;
 
    /* function for Kth element */
    function findK(A,i,j,n,m,k)
    {
        if (n < 1 || m < 1)
            return -1;
 
        /*.....If element is in outermost ring ....*/
        /* Element is in first row */
        if (k <= m)
            return A[i + 0][j + k - 1];
 
        /* Element is in last column */
        if (k <= (m + n - 1))
            return A[i + (k - m)][j + m - 1];
 
        /* Element is in last row */
        if (k <= (m + n - 1 + m - 1))
            return A[i + n - 1][j + m - 1 - (k - (m + n - 1))];
 
        /* Element is in first column */
        if (k <= (m + n - 1 + m - 1 + n - 2))
            return A[i + n - 1 - (k - (m + n - 1 + m - 1))][j + 0];
 
        /*.....If element is NOT in outermost ring ....*/
        /* Recursion for sub-matrix. &A[1][1] is
    address to next inside sub matrix.*/
        return findK(A, i + 1, j + 1, n - 2,
                    m - 2, k - (2 * n + 2 * m - 4));
    }
 
    /* Driver code */
     
        let a = [[ 1, 2, 3, 4, 5, 6 ],
                    [ 7, 8, 9, 10, 11, 12 ],
                    [ 13, 14, 15, 16, 17, 18 ]];
        let k = 17;
        document.write(findK(a, 0, 0, 3, 6, k));
     
// This code is contributed by sravan kumar
 

输出：

10

复杂性分析：

• 时间复杂度： O(c)，其中 c 是相对于第 k 个元素的外圆环数。
• 空间复杂度： O(1)。
  因为需要恒定的空间。