给定一个维度为N * M的矩阵arr[][]和一个整数K ，任务是以螺旋形式打印从左上角元素到K的矩阵的所有元素。
例子：
Input : N=5, M=6, K=15, arr[][]={{1, 2, 3, 4, 5, 6},
{7, 8, 9, 10, 11, 12},
{13, 14, 15, 16, 17, 18},
{19, 20, 21, 22, 23, 24},
{25, 26, 27, 28, 29, 30}}
Output: 1, 2, 7, 13, 8, 3, 4, 9, 14, 19, 25, 20, 15
Explanation:
| 1 | 2 | 3 | 4 | 5 | 6 |
| 7 | 8 | 9 | 10 | 11 | 12 |
| 13 | 14 | 15 | 16 | 17 | 18 |
| 19 | 20 | 21 | 22 | 23 | 24 |
| 25 | 26 | 27 | 28 | 29 | 30 |
1st diagonal printed: {1}
2nd diagonal printed: {2, 7}
3rd diagonal printed: {13, 8, 3}
……
5th diagonal printed {25, 20, 15}.
Since 15 is encountered, no further matrix element is printed.
Input: N = 4, M = 3, K = 69, arr[][]={{4, 87, 24},
{17, 1, 18},
{25, 69, 97},
{19, 27, 85}}
Output: 4, 87, 17, 25, 1, 24, 18, 69
处理方法：按照以下步骤解决问题：
- 这 矩阵中对角线的总数为N + M – 1 。
- 以螺旋方式一条一条地穿过对角线。
- 对于遍历的每个元素，检查它是否等于K。如果发现为真，则打印该元素并终止。
- 否则，打印元素并评估要遍历的下一个索引。如果i和j是当前索引：
- 沿对角线向上移动时， i将递减， j将递增。
- 沿对角线向下移动时， i将递增， j将递减。
- 如果下一个索引不是有效索引，则移动到下一个对角线。
- 否则，将当前位置更新到下一个。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if the
// indices are valid or not
bool isValid(int i, int j,
int N, int M)
{
return (i >= 0 && i < N
&& j >= 0 && j < M);
}
// Function to evaluate the next
// index while moving diagonally up
pair up(int i, int j,
int N, int M)
{
if (isValid(i - 1, j + 1, N, M))
return { i - 1, j + 1 };
else
return { -1, -1 };
}
// Function to evaluate the next
// index while moving diagonally down
pair down(int i, int j,
int N, int M)
{
if (isValid(i + 1, j - 1, N, M))
return { i + 1, j - 1 };
else
return { -1, -1 };
}
// Function to print matrix elements
// diagonally in Spiral Form
void SpiralDiagonal(int N, int M, int K,
vector > a)
{
int i = 0, j = 0;
// Total Number of Diagonals
// = N + M - 1
for (int diagonal = 0;
diagonal < N + M - 1;
diagonal++) {
while (1) {
// Stop when K is
// encountered
if (a[i][j] == K) {
cout << K;
return;
}
// Print the integer
cout << a[i][j] << ", ";
// Store the next index
pair next;
if (diagonal & 1) {
next = down(i, j, N, M);
}
else {
next = up(i, j, N, M);
}
// If current index is invalid
if (next.first == next.second
&& next.first == -1) {
// Move to the next diagonal
if (diagonal & 1) {
(i + 1 < N) ? ++i : ++j;
}
else {
(j + 1 < M) ? ++j : ++i;
}
break;
}
// Otherwise move to the
// next valid index
else {
i = next.first;
j = next.second;
}
}
}
}
// Driver Code
int main()
{
int N = 5, M = 6, K = 15;
vector > arr
= { { 1, 2, 3, 4, 5, 6 },
{ 7, 8, 9, 10, 11, 12 },
{ 13, 14, 15, 16, 17, 18 },
{ 19, 20, 21, 22, 23, 24 },
{ 25, 26, 27, 28, 29, 30 } };
// Function Call
SpiralDiagonal(N, M, K, arr);
return 0;
} Java
// Java program for the
// above approach
import java.util.*;
import java.lang.*;
class GFG{
static class pair
{
int first, second;
pair(int f, int s)
{
this.first = f;
this.second = s;
}
}
// Function to check if the
// indices are valid or not
static boolean isValid(int i, int j,
int N, int M)
{
return (i >= 0 && i < N &&
j >= 0 && j < M);
}
// Function to evaluate the next
// index while moving diagonally up
static int[] up(int i, int j,
int N, int M)
{
if (isValid(i - 1, j + 1, N, M))
return new int[]{ i - 1, j + 1 };
else
return new int[]{ -1, -1 };
}
// Function to evaluate the next
// index while moving diagonally down
static int[] down(int i, int j,
int N, int M)
{
if (isValid(i + 1, j - 1, N, M))
return new int[]{ i + 1, j - 1 };
else
return new int[]{ -1, -1 };
}
// Function to print matrix elements
// diagonally in Spiral Form
static void SpiralDiagonal(int N, int M,
int K, int[][] a)
{
int i = 0, j = 0;
// Total Number of Diagonals
// = N + M - 1
for(int diagonal = 0;
diagonal < N + M - 1;
diagonal++)
{
while (true)
{
// Stop when K is
// encountered
if (a[i][j] == K)
{
System.out.print(K);
return;
}
// Print the integer
System.out.print(a[i][j] + ", ");
// Store the next index
int[] next;
if ((diagonal & 1) == 1)
{
next = down(i, j, N, M);
}
else
{
next = up(i, j, N, M);
}
// If current index is invalid
if (next[0] == next[1] &&
next[1] == -1)
{
// Move to the next diagonal
if ((diagonal & 1) == 1)
{
if (i + 1 < N)
++i;
else
++j;
}
else
{
if (j + 1 < M)
++j;
else
++i;
}
break;
}
// Otherwise move to the
// next valid index
else
{
i = next[0];
j = next[1];
}
}
}
}
// Driver code
public static void main (String[] args)
{
int N = 5, M = 6, K = 15;
int[][] arr = { { 1, 2, 3, 4, 5, 6 },
{ 7, 8, 9, 10, 11, 12 },
{ 13, 14, 15, 16, 17, 18 },
{ 19, 20, 21, 22, 23, 24 },
{ 25, 26, 27, 28, 29, 30 } };
// Function Call
SpiralDiagonal(N, M, K, arr);
}
}
// This code is contributed by offbeatPython3
# Python3 program for the
# above approach
# Function to check if the
# indices are valid or not
def isValid(i, j, N, M):
return (i >= 0 and i < N and j >= 0 and j < M)
# Function to evaluate the next
# index while moving diagonally up
def up(i, j, N, M):
if(isValid(i - 1, j + 1, N, M)):
return [i - 1, j + 1 ]
else:
return [-1, -1]
# Function to evaluate the next
# index while moving diagonally down
def down(i, j, N, M):
if(isValid(i + 1, j - 1, N, M)):
return [i + 1, j - 1 ]
else:
return [-1, -1]
# Function to print matrix elements
# diagonally in Spiral Form
def SpiralDiagonal(N, M, K, a):
i = 0
j = 0
# Total Number of Diagonals
# = N + M - 1
for diagonal in range(N + M - 1):
while(True):
# Stop when K is
# encountered
if(a[i][j] == K):
print(K, end = "")
return
# Print the integer
print(a[i][j], ", ", end="", sep="")
# Store the next index
next = []
if((diagonal & 1) == 1):
next = down(i, j, N, M)
else:
next = up(i, j, N, M)
# If current index is invalid
if(next[0] == next[1] and next[1] == -1):
# Move to the next diagonal
if((diagonal & 1) == 1):
if(i + 1 < N):
i += 1
else:
j += 1
else:
if(j + 1 < M):
j += 1
else:
i += 1
break
# Otherwise move to the
# next valid index
else:
i = next[0]
j = next[1]
# Driver code
N = 5
M = 6
K = 15
arr = [[1, 2, 3, 4, 5, 6 ],
[ 7, 8, 9, 10, 11, 12],
[13, 14, 15, 16, 17, 18 ],
[19, 20, 21, 22, 23, 24],
[25, 26, 27, 28, 29, 30]]
# Function Call
SpiralDiagonal(N, M, K, arr);
# This code is contributed by avanitrachhadiya2155C#
// C# program for the
// above approach
using System;
class GFG{
// Function to check if the
// indices are valid or not
static bool isValid(int i, int j,
int N, int M)
{
return (i >= 0 && i < N &&
j >= 0 && j < M);
}
// Function to evaluate the next
// index while moving diagonally up
static int[] up(int i, int j, int N, int M)
{
if (isValid(i - 1, j + 1, N, M))
return new int[]{ i - 1, j + 1 };
else
return new int[]{ -1, -1 };
}
// Function to evaluate the next
// index while moving diagonally down
static int[] down(int i, int j, int N, int M)
{
if (isValid(i + 1, j - 1, N, M))
return new int[]{ i + 1, j - 1 };
else
return new int[]{ -1, -1 };
}
// Function to print matrix elements
// diagonally in Spiral Form
static void SpiralDiagonal(int N, int M, int K,
int[, ] a)
{
int i = 0, j = 0;
// Total Number of Diagonals
// = N + M - 1
for(int diagonal = 0;
diagonal < N + M - 1;
diagonal++)
{
while (true)
{
// Stop when K is
// encountered
if (a[i, j] == K)
{
Console.Write(K);
return;
}
// Print the integer
Console.Write(a[i, j] + ", ");
// Store the next index
int[] next;
if ((diagonal & 1) == 1)
{
next = down(i, j, N, M);
}
else
{
next = up(i, j, N, M);
}
// If current index is invalid
if (next[0] == next[1] &&
next[1] == -1)
{
// Move to the next diagonal
if ((diagonal & 1) == 1)
{
if (i + 1 < N)
++i;
else
++j;
}
else
{
if (j + 1 < M)
++j;
else
++i;
}
break;
}
// Otherwise move to the
// next valid index
else
{
i = next[0];
j = next[1];
}
}
}
}
// Driver code
public static void Main(string[] args)
{
int N = 5, M = 6, K = 15;
int[, ] arr = { { 1, 2, 3, 4, 5, 6 },
{ 7, 8, 9, 10, 11, 12 },
{ 13, 14, 15, 16, 17, 18 },
{ 19, 20, 21, 22, 23, 24 },
{ 25, 26, 27, 28, 29, 30 } };
// Function Call
SpiralDiagonal(N, M, K, arr);
}
}
// This code is contributed by grand_master输出：
1, 2, 7, 13, 8, 3, 4, 9, 14, 19, 25, 20, 15时间复杂度： O(N*M)
辅助空间： O(1)
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