通过用其左或右对角线和的最大值替换每个元素来修改矩阵

给定一个维度为M * N的矩阵 mat[][] ，任务是用其左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.

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)

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)

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