原文:
给定向量 arr[] 的
示例:
输入: arr[][] = {{1，2，3}，{4，5，6}，{7，8，9}} 输出: 1 4 2 7 5 3 8 6 9 解释: 以下是如何按对角线向上的顺序打印的图示:
输入: arr[][] = {{1，2，3，4，5}，{6，7}，{8}，{9，10，11}，{12，13，14，15，16}} 输出:1 6 28 7 3 9 4 12 10 5 13 11 14 15 16 解释: 下图是如何按对角线向上的顺序打印:
方法:这个想法是基于这样的观察:特定向上对角线中的所有元素的总和等于(行索引 列索引)。按照以下步骤解决问题:
- 初始化向量的向量 v ，以所需的格式存储元素。
- arr【】【】【】t3】，对于每个 i 和 j 将arr【i】【j】推至v【i j】。
- 完成上述步骤后，在 v 中。
- 现在，逐行打印存储在 v 中的所有元素，以获得所需的结果。
下面是上述方法的实现:
c
// c program for the above approach
#include
using namespace std;
// function to traverse the matrix
// diagonally upwards
void printdiagonaltraversal(
vector >& nums)
{
// stores the maximum size of vector
// from all row of matrix nums[][]
int max_size = nums.size();
for (int i = 0; i < nums.size(); i ) {
if (max_size < nums[i].size()) {
max_size = nums[i].size();
}
}
// store elements in desired order
vector > v(2 * max_size - 1);
// store every element on the basis
// of sum of index (i j)
for (int i = 0; i < nums.size(); i ) {
for (int j = 0;
j < nums[i].size(); j ) {
v[i j].push_back(nums[i][j]);
}
}
// print the stored result
for (int i = 0; i < v.size(); i ) {
// reverse all sublist
reverse(v[i].begin(), v[i].end());
for (int j = 0; j < v[i].size(); j )
cout << v[i][j] << " ";
}
}
// driver code
int main()
{
// given vector of vectors arr
vector > arr
= { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
// function call
printdiagonaltraversal(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.*;
class gfg{
// function to traverse the matrix
// diagonally upwards
static void printdiagonaltraversal(int[][] nums)
{
// stores the maximum size of vector
// from all row of matrix nums[][]
int max_size = nums[0].length;
// store elements in desired order
arraylist<
arraylist> v = new arraylist<
arraylist>();
for(int i = 0; i < 2 * max_size - 1; i )
{
v.add(new arraylist());
}
// store every element on the basis
// of sum of index (i j)
for(int i = 0; i < nums[0].length; i )
{
for(int j = 0; j < nums[0].length; j )
{
v.get(i j).add(nums[i][j]);
}
}
// print the stored result
for(int i = 0; i < v.size(); i )
{
// print in reverse order
for(int j = v.get(i).size() - 1;
j >= 0; j--)
{
system.out.print(v.get(i).get(j) " ");
}
}
}
// driver code
public static void main(string[] args)
{
// given vector of vectors arr
int[][] arr = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
// function call
printdiagonaltraversal(arr);
}
}
// this code is contributed by divyeshrabadiya07
python 3
# python3 program for the above approach
# function to traverse the matrix
# diagonally upwards
def printdiagonaltraversal(nums):
# stores the maximum size of vector
# from all row of matrix nums[][]
max_size = len(nums)
for i in range(len(nums)):
if (max_size < len(nums[i])):
max_size = len(nums[i])
# store elements in desired order
v = [[] for i in range(2 * max_size - 1)]
# store every element on the basis
# of sum of index (i j)
for i in range(len(nums)):
for j in range(len(nums[i])):
v[i j].append(nums[i][j])
# print the stored result
for i in range(len(v)):
# reverse all sublist
v[i] = v[i][::-1]
for j in range(len(v[i])):
print(v[i][j], end = " ")
# driver code
if __name__ == '__main__':
# given vector of vectors arr
arr = [ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ] ]
# function call
printdiagonaltraversal(arr)
# this code is contributed by mohit kumar 29
c
// c# program for the above approach
using system;
using system.collections.generic;
class gfg{
// function to traverse the matrix
// diagonally upwards
static void printdiagonaltraversal(int[,] nums)
{
// stores the maximum size of vector
// from all row of matrix nums[][]
int max_size = nums.getlength(0);
// store elements in desired order
list> v = new list>();
for(int i = 0; i < 2 * max_size - 1; i )
{
v.add(new list());
}
// store every element on the basis
// of sum of index (i j)
for(int i = 0; i < nums.getlength(0); i )
{
for(int j = 0; j < nums.getlength(0); j )
{
v[i j].add(nums[i, j]);
}
}
// print the stored result
for(int i = 0; i < v.count; i )
{
// print in reverse order
for(int j = v[i].count - 1; j >= 0; j--)
{
console.write(v[i][j] " ");
}
}
}
// driver code
static void main()
{
// given vector of vectors arr
int[,] arr = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
// function call
printdiagonaltraversal(arr);
}
}
// this code is contributed by divyesh072019
java 描述语言
output:
1 4 2 7 5 3 8 6 9
时间复杂度: o(nm)，其中 n 是给定矩阵的大小，m 是矩阵中任意一行的最大大小。* 辅助空间: o(nm)*
交替进场:使用也可以解决上述问题。按照以下步骤解决问题:
- 初始化一个队列 q ，插入 arr[][] 第一个单元格的索引，即 (0，0) 。
- 初始化一个 v ，以所需的格式存储元素。
- 当 q 不为空时，请执行以下操作:
- 前面的元素，推入 v 。
- 仅当当前单元格是其所在行的第一个单元格时，将当前单元格的索引推到其正下方。
- 如果其右邻小区存在，则推送其索引。
- 完成上述步骤后，打印存储在 v 中的所有元素。
下面是上述方法的实现:
c
// c program for the above approach
#include
using namespace std;
// function to traverse the matrix
// diagonally upwards
void printdiagonaltraversal(
vector >& nums)
{
// store the number of rows
int m = nums.size();
// initialize queue
queue > q;
// push the index of first element
// i.e., (0, 0)
q.push({ 0, 0 });
while (!q.empty()) {
// get the front element
pair p = q.front();
// pop the element at the front
q.pop();
cout << nums[p.first][p.second]
<< " ";
// insert the element below
// if the current element is
// in first column
if (p.second == 0
&& p.first 1 < m) {
q.push({ p.first 1,
p.second });
}
// insert the right neighbour
// if it exists
if (p.second 1 < nums[p.first].size())
q.push({ p.first,
p.second 1 });
}
}
// driver code
int main()
{
// given vector of vectors arr
vector > arr
= { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
// function call
printdiagonaltraversal(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.*;
class gfg{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// function to traverse the matrix
// diagonally upwards
static void printdiagonaltraversal(
int [][]nums)
{
// store the number of rows
int m = nums.length;
// initialize queue
queue q = new linkedlist<>();
// push the index of first element
// i.e., (0, 0)
q.add(new pair( 0, 0 ));
while (!q.isempty()) {
// get the front element
pair p = q.peek();
// pop the element at the front
q.remove();
system.out.print(nums[p.first][p.second]
" ");
// insert the element below
// if the current element is
// in first column
if (p.second == 0
&& p.first 1 < m) {
q.add(new pair( p.first 1,
p.second ));
}
// insert the right neighbour
// if it exists
if (p.second 1 < nums[p.first].length)
q.add(new pair( p.first,
p.second 1 ));
}
}
// driver code
public static void main(string[] args)
{
// given vector of vectors arr
int[][] arr
= { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
// function call
printdiagonaltraversal(arr);
}
}
// this code is contributed by amit katiyar
python 3
# python3 program for the above approach
# function to traverse the matrix
# diagonally upwards
def printdiagonaltraversal(nums):
# store the number of rows
m = len(nums)
# initialize queue
q = []
# push the index of first element
# i.e., (0, 0)
q.append([ 0, 0 ])
while (len(q) != 0):
# get the front element
p = q[0]
# pop the element at the front
q.pop(0);
print(nums[p[0]][p[1]], end = " ")
# insert the element below
# if the current element is
# in first column
if (p[1] == 0
and p[0] 1 < m):
q.append([ p[0] 1,
p[1] ]);
# insert the right neighbour
# if it exists
if (p[1] 1 < len(nums[p[0]])):
q.append([ p[0],
p[1] 1 ]);
# driver code
if __name__ == "__main__":
# given vector of vectors arr
arr = [[ 1, 2, 3 ], [ 4, 5, 6 ] ,[ 7, 8, 9 ]]
# function call
printdiagonaltraversal(arr);
# this code is contributed by chitranayal
c
// c# program for the above approach
using system;
using system.collections.generic;
public class gfg
{
class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// function to traverse the matrix
// diagonally upwards
static void printdiagonaltraversal(
int [,]nums)
{
// store the number of rows
int m = nums.getlength(0);
// initialize queue
queue q = new queue();
// push the index of first element
// i.e., (0, 0)
q.enqueue(new pair(0, 0));
while (q.count != 0)
{
// get the front element
pair p = q.peek();
// pop the element at the front
q.dequeue();
console.write(nums[p.first,p.second]
" ");
// insert the element below
// if the current element is
// in first column
if (p.second == 0
&& p.first 1 < m)
{
q.enqueue(new pair( p.first 1,
p.second ));
}
// insert the right neighbour
// if it exists
if (p.second 1 < nums.getlength(1))
q.enqueue(new pair( p.first,
p.second 1 ));
}
}
// driver code
public static void main(string[] args)
{
// given vector of vectors arr
int[,] arr
= { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
// function call
printdiagonaltraversal(arr);
}
}
// this code is contributed by shikhasingrajput
java 描述语言
output:
1 4 2 7 5 3 8 6 9
时间复杂度: o(nm)，其中 n 是给定矩阵的大小，m 是矩阵中任意一行的最大大小。* 辅助空间: o(n)
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