给定一个大小为N × N的矩阵mat[][] ，任务是最小化相邻行交换的计数，以将给定的矩阵转换为下三角矩阵。如果无法将给定矩阵转换为下三角矩阵，则打印 -1。
注意：下三角矩阵在主对角线上方的所有索引处都包含 0。
例子：
Input: mat[][] = {{0, 0, 2}, {3, 1, 0}, {4, 0, 0}}
Output: 3
Explanation:
Swap 1st row and 2nd row {{3, 1, 0}, {0, 0, 2}, {4, 0, 0}}
Swap 2nd row and 3rd row {{3, 1, 0}, {4, 0, 0}, {0, 0, 2}}
Swap 1st row and 2nd row {{4, 0, 0}, {3, 1, 0}, {0, 0, 2}}
Therefore, the required output is 3.
Input: mat[][] = {{0, 2, 3, 0}, {0, 2, 4, 0}, {0, 5, 1, 0}, {0, 1, 1, 0}}
Output: -1
方法：这个问题可以使用贪心技术来解决。首先找到每行中存在的零数并将其存储在整数数组中的想法。然后，根据0的数量降序计算对数组进行排序所需的最小相邻交换数。请按照以下步骤解决问题：
- 初始化一个数组，比如cntZero[]来存储每行中存在的0的计数。
- 对于第i行，从第(i + 1)个索引遍历cntZero[]数组并找到第一个索引，比如First where ctnZero[First] >= (N – i -1) 。
- 将所有相邻元素从第i个索引交换到cntZero[]数组的第一个索引并增加计数。
- 最后，返回将给定矩阵转换为下三角矩阵所需的相邻交换计数。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to count the minimum
// number of adjacent swaps
int minAdjSwaps(vector >& mat)
{
// Stores the size of
// the given matrix
int N = mat.size();
// Stores the count of zero
// at the end of each row
vector cntZero(N, 0);
// Traverse the given matrix
for (int i = 0; i < N; i++) {
// Count of 0s at the end
// of the ith row
for (int j = N - 1;
j >= 0 && mat[i][j] == 0;
j--) {
cntZero[i]++;
}
}
// Stores the count of swaps
int cntSwaps = 0;
// Traverse the cntZero array
for (int i = 0; i < N; i++) {
// If count of zero in the
// i-th row < (N - i - 1)
if (cntZero[i]
< (N - i - 1)) {
// Stores the index of the row
// where count of zero > (N-i-1)
int First = i;
while (First < N
&& cntZero[First]
< (N - i - 1)) {
First++;
}
// If no row found that
// satisfy the condition
if (First == N) {
return -1;
}
// Swap the adjacent row
while (First > i) {
swap(cntZero[First],
cntZero[First - 1]);
First--;
cntSwaps++;
}
}
}
return cntSwaps;
}
// Driver Code
int main()
{
vector > mat
= { { 0, 0, 2 },
{ 3, 1, 0 },
{ 4, 0, 0 } };
cout << minAdjSwaps(mat);
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to count the minimum
// number of adjacent swaps
static int minAdjSwaps(int [][]mat)
{
// Stores the size of
// the given matrix
int N = mat.length;
// Stores the count of zero
// at the end of each row
int []cntZero = new int[N];
// Traverse the given matrix
for(int i = 0; i < N; i++)
{
// Count of 0s at the end
// of the ith row
for(int j = N - 1;
j >= 0 && mat[i][j] == 0;
j--)
{
cntZero[i]++;
}
}
// Stores the count of swaps
int cntSwaps = 0;
// Traverse the cntZero array
for(int i = 0; i < N; i++)
{
// If count of zero in the
// i-th row < (N - i - 1)
if (cntZero[i] < (N - i - 1))
{
// Stores the index of the row
// where count of zero > (N-i-1)
int First = i;
while (First < N && cntZero[First] <
(N - i - 1))
{
First++;
}
// If no row found that
// satisfy the condition
if (First == N)
{
return -1;
}
// Swap the adjacent row
while (First > i)
{
cntZero = swap(cntZero, First,
First - 1);
First--;
cntSwaps++;
}
}
}
return cntSwaps;
}
static int[] swap(int []arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Driver Code
public static void main(String[] args)
{
int [][]mat = { { 0, 0, 2 },
{ 3, 1, 0 },
{ 4, 0, 0 } };
System.out.print(minAdjSwaps(mat));
}
}
// This code is contributed by Amit KatiyarPython3
# Python3 program to implement
# the above approach
# Function to count the minimum
# number of adjacent swaps
def minAdjSwaps(mat):
# Stores the size of
# the given matrix
N = len(mat)
# Stores the count of zero
# at the end of each row
cntZero = [0] * (N)
# Traverse the given matrix
for i in range(N):
# Count of 0s at the end
# of the ith row
for j in range(N - 1, -1, -1):
if mat[i][j] != 0:
break
cntZero[i] += 1
# Stores the count of swaps
cntSwaps = 0
# Traverse the cntZero array
for i in range(N):
# If count of zero in the
# i-th row < (N - i - 1)
if (cntZero[i] < (N - i - 1)):
# Stores the index of the row
# where count of zero > (N-i-1)
First = i
while (First < N and
cntZero[First] < (N - i - 1)):
First += 1
# If no row found that
# satisfy the condition
if (First == N):
return -1
# Swap the adjacent row
while (First > i):
cntZero[First] = cntZero[First - 1]
cntZero[First - 1] = cntZero[First]
First -= 1
cntSwaps += 1
return cntSwaps
# Driver Code
if __name__ == '__main__':
mat = [ [ 0, 0, 2 ],
[ 3, 1, 0 ],
[ 4, 0, 0 ] ]
print(minAdjSwaps(mat))
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to count the minimum
// number of adjacent swaps
static int minAdjSwaps(int [,]mat)
{
// Stores the size of
// the given matrix
int N = mat.GetLength(0);
// Stores the count of zero
// at the end of each row
int []cntZero = new int[N];
// Traverse the given matrix
for(int i = 0; i < N; i++)
{
// Count of 0s at the end
// of the ith row
for(int j = N - 1;
j >= 0 && mat[i, j] == 0;
j--)
{
cntZero[i]++;
}
}
// Stores the count of swaps
int cntSwaps = 0;
// Traverse the cntZero array
for(int i = 0; i < N; i++)
{
// If count of zero in the
// i-th row < (N - i - 1)
if (cntZero[i] < (N - i - 1))
{
// Stores the index of the row
// where count of zero > (N-i-1)
int First = i;
while (First < N &&
cntZero[First] < (N - i - 1))
{
First++;
}
// If no row found that
// satisfy the condition
if (First == N)
{
return -1;
}
// Swap the adjacent row
while (First > i)
{
cntZero = swap(cntZero,
First, First - 1);
First--;
cntSwaps++;
}
}
}
return cntSwaps;
}
static int[] swap(int []arr,
int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
// Driver Code
public static void Main(String[] args)
{
int [,]mat = {{0, 0, 2},
{3, 1, 0},
{4, 0, 0}};
Console.Write(minAdjSwaps(mat));
}
}
// This code is contributed by 29AjayKumarJavascript
输出
3时间复杂度： O(N 2 )
辅助空间： O(N)
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