原文:
给定一个尺寸为 n * m 的arr【】【】和一个整数 k ，任务是以螺旋形式对角打印矩阵中从左上角元素到 k 的所有元素。
示例:
输入: 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}，
| one | two | three | four | five | six | | seven | eight | nine | ten | eleven | twelve | | thirteen | fourteen | fifteen | sixteen | seventeen | eighteen | | nineteen | twenty | twenty-one | twenty-two | twenty-three | twenty-four | | twenty-five | twenty-six | twenty-seven | twenty-eight | twenty-nine | thirty |
1 st 对角印花:{1} 2 nd 对角印花:{2，7} 3 rd 对角印花:{13，8，3} …… 5 th 对角印花{25，20，15}。 由于遇到 15，因此不再打印矩阵元素。
输入: n = 4，m = 3，k = 69，arr[][]={{4，87，24}， {17，1，18}， {25，69，97}， {19，27，85}} 输出: 4，87，17，25，1，24，18，69
方法:按照以下步骤解决问题:
- 矩阵中的对角线总数为n m–1。
- 以螺旋方式逐个穿过对角线。
- 对于遍历的每个元素，检查是否等于 k 。如果发现是真的，打印该元素并终止。
- 否则，打印元素并计算要遍历的下一个索引。如果 i 和 j 是当前指数:
- 沿对角线向上移动时，i 将减少，而 j 将增加。
- 对角向下移动时，i 将增加，而 j 将减少。
- 如果下一个索引不是有效的索引，则移动到下一个对角线。
- 否则，将当前位置更新到下一个位置。
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 语言(一种计算机语言，尤用于创建网站)
// 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 offbeat
python 3
# 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 avanitrachhadiya2155
c
// 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
java 描述语言
输出:
1, 2, 7, 13, 8, 3, 4, 9, 14, 19, 25, 20, 15
时间复杂度: o(nm)* 辅助空间: o(1)
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