给定一个维度为M * N的矩阵mat[][] ,任务是用其左对角线或右对角线的最大和替换每个矩阵元素。
例子:
Input: mat[][] = {{5, 2, 1}, {7, 2, 6}, {3, 1, 9}}
Output:
16 9 6
9 16 8
6 8 16
Explanation:
Replace each element with max(sum of right diagonal, sum of left diagonal).
Follow the diagram below to understand more clearly.
Input: mat[][] = {{1, 2}, {3, 4}}
Output:
5 5
5 5
方法:主要思想基于以下观察的事实:
- 右对角线元素的行索引和列索引之和相等。
- 左对角元素的行索引和列索引之间的差值相等。
- 使用上面的两个属性,使用一个Map来存储每个元素左右对角线的总和。
- 遍历矩阵并用左对角线和或右对角线和的最大值替换每个元素。
- 打印得到的最终矩阵。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to update given matrix with
// maximum of left and right diagonal sum
void updateMatrix(int mat[][3])
{
// Stores the total sum
// of right diagonal
map right;
// Stores the total sum
// of left diagonal
map left;
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
// Update the map storing
// right diagonal sums
if (right.find(i + j) == right.end())
right[i + j] = mat[i][j];
else
right[i + j] += mat[i][j];
// Update the map storing
// left diagonal sums
if (left.find(i - j) == left.end())
left[i - j] = mat[i][j];
else
left[i - j] += mat[i][j];
}
}
// Traverse the matrix
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
// Update the matrix
mat[i][j] = max(right[i + j], left[i - j]);
}
}
// Print the matrix
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
cout << mat[i][j] << " ";
}
cout << endl;
}
}
// Driver code
int main()
{
int mat[][3]
= { { 5, 2, 1 }, { 7, 2, 6 }, { 3, 1, 9 } };
updateMatrix(mat);
return 0;
}
// This code is contributed by ukasp.
Java
// Java program for the above approach
// Function to update given matrix with
// maximum of left and right diagonal sum
import java.io.*;
import java.util.*;
class GFG {
static void updateMatrix(int mat[][])
{
Map right
= new HashMap();
Map left
= new HashMap();
for(int i = 0; i < 3; i++)
{
for(int j = 0; j < 3; j++)
{
// Update the map storing
// right diagonal sums
if (!right.containsKey(i + j))
right.put(i + j, mat[i][j]);
else
right.put(i + j,
right.get(i + j) + mat[i][j]);
// Update the map storing
// left diagonal sums
if (!left.containsKey(i - j))
left.put(i - j, mat[i][j]);
else
left.put(i - j,
left.get(i - j) + mat[i][j]);
}
}
// Traverse the matrix
for(int i = 0; i < 3; i++)
{
for(int j = 0; j < 3; j++)
{
// Update the matrix
mat[i][j] = Math.max(right.get(i + j),
left.get(i - j));
}
}
// Print the matrix
for(int i = 0; i < 3; i++)
{
for(int j = 0; j < 3; j++)
{
System.out.print(mat[i][j] + " ");
}
System.out.print("\n");
}
}
// Driver code
public static void main (String[] args) {
int[][] mat = {{ 5, 2, 1 },
{ 7, 2, 6 },
{ 3, 1, 9 }};
updateMatrix(mat);
}
}
// This code is contributed by avanitrachhadiya2155
Python3
# Python3 program for the above approach
# Function to update given matrix with
# maximum of left and right diagonal sum
def updateMatrix(mat):
# Stores the total sum
# of right diagonal
right = {}
# Stores the total sum
# of left diagonal
left = {}
for i in range(len(mat)):
for j in range(len(mat[0])):
# Update the map storing
# right diagonal sums
if i + j not in right:
right[i + j] = mat[i][j]
else:
right[i + j] += mat[i][j]
# Update the map storing
# left diagonal sums
if i-j not in left:
left[i-j] = mat[i][j]
else:
left[i-j] += mat[i][j]
# Traverse the matrix
for i in range(len(mat)):
for j in range(len(mat[0])):
# Update the matrix
mat[i][j] = max(right[i + j], left[i-j])
# Print the matrix
for i in mat:
print(*i)
# Given matrix
mat = [[5, 2, 1], [7, 2, 6], [3, 1, 9]]
updateMatrix(mat)
C#
// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
// Function to update given matrix with
// maximum of left and right diagonal sum
static void updateMatrix(int[,] mat)
{
// Stores the total sum
// of right diagonal
Dictionary right = new Dictionary();
// Stores the total sum
// of left diagonal
Dictionary left = new Dictionary();
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
// Update the map storing
// right diagonal sums
if (!right.ContainsKey(i + j))
right[i + j] = mat[i,j];
else
right[i + j] += mat[i,j];
// Update the map storing
// left diagonal sums
if (!left.ContainsKey(i - j))
left[i - j] = mat[i,j];
else
left[i - j] += mat[i,j];
}
}
// Traverse the matrix
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
// Update the matrix
mat[i,j] = Math.Max(right[i + j], left[i - j]);
}
}
// Print the matrix
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
Console.Write(mat[i,j] + " ");
}
Console.WriteLine();
}
}
static void Main ()
{
int[,] mat = { { 5, 2, 1 }, { 7, 2, 6 }, { 3, 1, 9 } };
updateMatrix(mat);
}
}
// This code is contributed by suresh07.
Javascript
输出:
16 9 6
9 16 8
6 8 16
时间复杂度: O(N * M)
辅助空间: O(N * M)
如果您希望与专家一起参加现场课程,请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。