在二元矩阵中心得到 1 的最小步数
给定一个 N * N 矩阵，其中 N 与所有 0 值都是奇数，只有一个单元格的值为 1。任务是找到最小可能的移动，当在一次移动中，任何两个连续的行或两个连续的列可以交换时，将这个 1 移动到矩阵的中心。 例:
输入: mat[][] = { {0，0，1，0，0}、 {0，0，0，0，0}、 {0，0，0，0，0，0}、 {0，0，0，0，0，0}、 {0，0，0，0，0，0，0}} 输出: 2 操作 1:交换第一行和第二行。 操作 2:交换第二排和第三排。 输入: mat[][] = { {0，0，0}， {0，1，0}， {0，0，0}} 输出: 0
进场:
- 计算矩阵中心的位置为 (cI，cJ) = (⌊N / 2⌋，⌊N / 2⌋) 。
- 找到 1 在矩阵中的位置，存储在 (oneI，oneJ) 中。
- 现在，最小可能的移动将是ABS(cI–OneI)+ABS(Cj–OneJ)。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
const int N = 5;
// Function to return the
// minimum moves required
int minMoves(int mat[N][N])
{
// Center of the matrix
int cI = N / 2, cJ = N / 2;
// Find the position of the 1
int oneI = 0, oneJ = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (mat[i][j] == 1) {
oneI = i;
oneJ = j;
break;
}
}
}
return (abs(cI - oneI) + abs(cJ - oneJ));
}
// Driver code
int main()
{
int mat[N][N] = { { 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0 },
{ 0, 0, 1, 0, 0 } };
cout << minMoves(mat);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
static int N = 5;
// Function to return the
// minimum moves required
static int minMoves(int mat[][])
{
// Center of the matrix
int cI = N / 2, cJ = N / 2;
// Find the position of the 1
int oneI = 0, oneJ = 0;
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
if (mat[i][j] == 1)
{
oneI = i;
oneJ = j;
break;
}
}
}
return (Math.abs(cI - oneI) + Math.abs(cJ - oneJ));
}
// Driver code
public static void main(String[] args)
{
int mat[][] = { { 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0 },
{ 0, 0, 1, 0, 0 } };
System.out.print(minMoves(mat));
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python3 implementation of the approach
N = 5
# Function to return the
# minimum moves required
def minMoves(mat):
# Center of the matrix
cI = N // 2
cJ = N // 2
# Find the position of the 1
oneI = 0
oneJ = 0
for i in range(N):
for j in range(N):
if (mat[i][j] == 1):
oneI = i
oneJ = j
break
return (abs(cI - oneI) + abs(cJ - oneJ))
# Driver code
mat = [[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 1, 0, 0]]
print(minMoves(mat))
# This code is contributed by Mohit Kumar
C
// C# implementation of the approach
using System;
class GFG
{
static int N = 5;
// Function to return the
// minimum moves required
static int minMoves(int [,]mat)
{
// Center of the matrix
int cI = N / 2, cJ = N / 2;
// Find the position of the 1
int oneI = 0, oneJ = 0;
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
if (mat[i, j] == 1)
{
oneI = i;
oneJ = j;
break;
}
}
}
return (Math.Abs(cI - oneI) +
Math.Abs(cJ - oneJ));
}
// Driver code
public static void Main(String[] args)
{
int [,]mat = {{ 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0 },
{ 0, 0, 1, 0, 0 }};
Console.Write(minMoves(mat));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// JavaScript implementation of the approach
const N = 5;
// Function to return the
// minimum moves required
function minMoves(mat)
{
// Center of the matrix
let cI = parseInt(N / 2), cJ = parseInt(N / 2);
// Find the position of the 1
let oneI = 0, oneJ = 0;
for (let i = 0; i < N; i++) {
for (let j = 0; j < N; j++) {
if (mat[i][j] == 1) {
oneI = i;
oneJ = j;
break;
}
}
}
return (Math.abs(cI - oneI) + Math.abs(cJ - oneJ));
}
// Driver code
let mat = [ [ 0, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0 ],
[ 0, 0, 1, 0, 0 ] ];
document.write(minMoves(mat));
</script>
Output:
2
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