给定尺寸为N * N的矩阵M [] [] ，该矩阵仅由1到N范围内的整数组成，任务是计算主对角线上存在的矩阵元素的总和，包含重复行和列的行数价值观。
例子：
Input: N = 4, M[][] = {{1, 2, 3, 4}, {2, 1, 4, 3}, {3, 4, 1, 2}, {4, 3, 2, 1}}
Output: 4 0 0
Explanation:
Sum of diagonal = M[0][0] + M[1][1] + M[2][2] + M[3][3] = 4.
No row or column consists of repeated elements.
Therefore, the required sum is 4
Input: N = 3, M[][] = {{2, 1, 3}, {1, 3, 2}, {1, 2, 3}}
Output: 8 0 2
Explanation:
Sum of diagonal = M[0][0]+M[1][1]+M[2][2] = 8.
No row consists of repeated elements.
1st and 3rd columns consists of repeated elements.
方法：该方法是简单地遍历矩阵的所有元素，并找到对角线的总和以及具有重复值的行和列的数量。请按照以下步骤解决给定的问题：
- 初始化变量trace ， rowRepeat和columnRepeat分别存储主对角线的总和，行数和包含重复矩阵元素的列。
- 遍历矩阵M [i] [j]中存在的每个元素，并在i等于j时增加总和迹线。
- 要查找包含重复值的行，请一次遍历一行，比较值，然后检查是否存在重复值。在获得第一对重复元素后，将rowRepeat递增1并退出循环。
- 对矩阵的每一行重复上述步骤。
- 对所有列重复相同的过程，如果值匹配，则columnRepeat会增加1 。
- 完成所有迭代后，将trace ， rowRepeat和columnRepeat的值打印为结果。
下面是上述方法的实现：
C++14
// C++14 program for the above approach
#include
using namespace std;
// Function to calculate trace of
// a matrix and number of rows and
// columns with repeated elements
void vestigium(int N, int M[4][4])
{
// Stores the trace, number of
// rows and columns consisting
// of repeated matrix elements
int trace = 0, row_repeat = 0;
int column_repeat = 0;
// Iterate over the matrix
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
// If current element is
// present in the main diagonal
if (i == j)
{
// Update trace
trace += M[i][j];
}
}
int flag1 = 0;
// Iterate over each row
// and increment row_repeat
// if repeated values exists
for(int j = 0; j < N; j++)
{
for(int k = 0; k < N; k++)
{
// For each valid range
if (j != k && M[i][j] == M[i][k])
{
row_repeat++;
flag1 = 1;
break;
}
}
if (flag1 == 1)
{
break;
}
}
int flag2 = 0;
// Iterate over each column and
// increment column_repeat if
// repeated values are encountered
for(int j = 0; j < N; j++)
{
for(int k = 0; k < N; k++)
{
// For each valid range
if (j != k && M[j][i] == M[k][i])
{
column_repeat++;
flag2 = 1;
break;
}
}
if (flag2 == 1)
{
break;
}
}
}
// Answer
cout << trace << " "
<< row_repeat << " "
<< column_repeat ;
}
// Driver Code
int main()
{
int M[4][4] = { { 1, 2, 3, 4 },
{ 2, 1, 4, 3 },
{ 3, 4, 1, 2 },
{ 4, 3, 2, 1 } };
int N = sizeof(M) / sizeof(M[0]);
vestigium(N, M);
}
// This code is contributed by sanjoy_62 Java
// Java program for the above approach
public class GFG {
// Function to calculate trace of
// a matrix and number of rows and
// columns with repeated elements
public static String vestigium(
int N, int M[][])
{
// Stores the trace, number of
// rows and columns consisting
// of repeated matrix elements
int trace = 0, row_repeat = 0;
int column_repeat = 0;
// Iterate over the matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
// If current element is
// present in the main diagonal
if (i == j) {
// Update trace
trace += M[i][j];
}
}
int flag1 = 0;
// Iterate over each row
// and increment row_repeat
// if repeated values exists
for (int j = 0; j < N; j++) {
for (int k = 0; k < N; k++) {
// For each valid range
if (j != k
&& M[i][j] == M[i][k]) {
row_repeat++;
flag1 = 1;
break;
}
}
if (flag1 == 1) {
break;
}
}
int flag2 = 0;
// Iterate over each column and
// increment column_repeat if
// repeated values are encountered
for (int j = 0; j < N; j++) {
for (int k = 0; k < N; k++) {
// For each valid range
if (j != k
&& M[j][i] == M[k][i]) {
column_repeat++;
flag2 = 1;
break;
}
}
if (flag2 == 1) {
break;
}
}
}
// Answer
String output = trace + " "
+ row_repeat + " "
+ column_repeat + "\n";
// Return answer
return output;
}
// Driver Code
public static void main(String[] args)
{
int M[][] = { { 1, 2, 3, 4 },
{ 2, 1, 4, 3 },
{ 3, 4, 1, 2 },
{ 4, 3, 2, 1 } };
int N = M.length;
String output = vestigium(N, M);
// Print the output
System.out.print(output);
}
}Python3
# Python3 program for the above approach
# Function to calculate trace of
# a matrix and number of rows and
# columns with repeated elements
def vestigium(N, M) :
# Stores the trace, number of
# rows and columns consisting
# of repeated matrix elements
trace = 0
row_repeat = 0
column_repeat = 0
# Iterate over the matrix
for i in range(N) :
for j in range(N) :
# If current element is
# present in the main diagonal
if (i == j):
# Update trace
trace += M[i][j]
flag1 = 0
# Iterate over each row
# and increment row_repeat
# if repeated values exists
for j in range(N) :
for k in range(N) :
# For each valid range
if (j != k and M[i][j] == M[i][k]) :
row_repeat += 1
flag1 = 1
break
if (flag1 == 1) :
break
flag2 = 0
# Iterate over each column and
# increment column_repeat if
# repeated values are encountered
for j in range(N) :
for k in range(N) :
# For each valid range
if (j != k and M[j][i] == M[k][i]) :
column_repeat += 1
flag2 = 1
break
if (flag2 == 1) :
break
# Answer
print(trace, row_repeat, column_repeat)
# Driver Code
M = [[ 1, 2, 3, 4 ],
[ 2, 1, 4, 3 ],
[ 3, 4, 1, 2 ],
[ 4, 3, 2, 1 ]]
N = len(M)
vestigium(N, M)
# This code is contributed by avijitmonald1998.C#
// C# program for the above approach
using System;
class GFG{
// Function to calculate trace of
// a matrix and number of rows and
// columns with repeated elements
public static String vestigium(int N, int[,] M)
{
// Stores the trace, number of
// rows and columns consisting
// of repeated matrix elements
int trace = 0, row_repeat = 0;
int column_repeat = 0;
// Iterate over the matrix
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
// If current element is
// present in the main diagonal
if (i == j)
{
// Update trace
trace += M[i, j];
}
}
int flag1 = 0;
// Iterate over each row
// and increment row_repeat
// if repeated values exists
for(int j = 0; j < N; j++)
{
for(int k = 0; k < N; k++)
{
// For each valid range
if (j != k && M[i, j] == M[i, k])
{
row_repeat++;
flag1 = 1;
break;
}
}
if (flag1 == 1)
{
break;
}
}
int flag2 = 0;
// Iterate over each column and
// increment column_repeat if
// repeated values are encountered
for(int j = 0; j < N; j++)
{
for(int k = 0; k < N; k++)
{
// For each valid range
if (j != k && M[j, i] == M[k, i])
{
column_repeat++;
flag2 = 1;
break;
}
}
if (flag2 == 1)
{
break;
}
}
}
// Answer
string output = trace + " " + row_repeat + " " +
column_repeat + "\n";
// Return answer
return output;
}
// Driver Code
public static void Main(string[] args)
{
int[, ] M = { { 1, 2, 3, 4 },
{ 2, 1, 4, 3 },
{ 3, 4, 1, 2 },
{ 4, 3, 2, 1 } };
int N = M.GetLength(0);
string output = vestigium(N, M);
// Print the output
Console.Write(output);
}
}
// This code is contributed by ukaspJavascript
输出：
4 0 0时间复杂度： O(N 3 )
辅助空间： O(N 2 )
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