通过用矩阵的左对角线和或右对角线和的最大值替换每个元素来修改矩阵
原文:https://www . geesforgeks . org/modify-a-matrix-by-将每个元素替换为其左对角线或右对角线总和的最大值/
给定一个矩阵 mat[][] ，其维数为 M * N ，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。
示例:
输入: mat[][] = {{5，2，1}，{7，2，6}，{3，1，9}} 输出: 16 9 6 9 16 8 6 8 16 解释: 将每个元素替换为 max(右对角线之和，左对角线之和)。 按照下图理解更清楚。
输入: mat[][] = {{1，2}，{3，4 } } T3】输出:T5】5 5 5 5
方法:主要思想是基于以下观察的事实:
- 右对角线元素的行和列索引之和相等。
- 左对角线元素的行索引和列索引之间的差异相等。
- 使用以上两个属性，使用映射存储每个元素的左右对角线之和。
- 遍历矩阵，用左对角线和或右对角线和的最大值替换每个元素。
- 打印获得的最终矩阵。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
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<int, int> right;
// Stores the total sum
// of left diagonal
map<int, int> 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 语言(一种计算机语言，尤用于创建网站)
// 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<Integer, Integer> right
= new HashMap<Integer, Integer>();
Map<Integer, Integer> left
= new HashMap<Integer, Integer>();
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
Python 3
# 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<int, int> right = new Dictionary<int, int>();
// Stores the total sum
// of left diagonal
Dictionary<int, int> left = new Dictionary<int, int>();
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.
java 描述语言
<script>
// Javascript program for the above approach
// Function to update given matrix with
// maximum of left and right diagonal sum
function updateMatrix(mat)
{
// Stores the total sum
// of right diagonal
let right = new Map();
// Stores the total sum
// of left diagonal
let left = new Map();
for(let i = 0; i < 3; i++)
{
for(let j = 0; j < 3; j++)
{
// Update the map storing
// right diagonal sums
if (!right.has(i + j))
right.set(i + j, mat[i][j]);
else
right.set(i + j,
right.get(i + j) + mat[i][j]);
// Update the map storing
// left diagonal sums
if (!left.has(i - j))
left.set(i - j, mat[i][j]);
else
left.set(i - j,
left.get(i - j) + mat[i][j]);
}
}
// Traverse the matrix
for(let i = 0; i < 3; i++)
{
for(let 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(let i = 0; i < 3; i++)
{
for(let j = 0; j < 3; j++)
{
document.write(mat[i][j] + " ");
}
document.write("<br>");
}
}
// Driver code
let mat = [ [ 5, 2, 1 ],
[ 7, 2, 6 ],
[ 3, 1, 9 ] ];
updateMatrix(mat);
// This code is contributed by gfgking.
</script>
Output:
16 9 6
9 16 8
6 8 16
时间复杂度: O(N * M) 辅助空间: O(N * M)
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