逆时针旋转矩阵M * N的每个环 [英] Rotate each ring of the matrix M*N anticlockwise
问题描述
我无法沿逆时针方向旋转 M * N
矩阵.我的代码对于 3 * 3
矩阵可以正常工作,但是当我尝试其他任何情况时都无法正常工作,请假设我正在为 4 * 4
矩阵进行操作,然后仅外部元素正在旋转,而内部4个元素(即6,7,10,11)没有旋转.我的输入是1-16个数字,作为 4 * 4
矩阵:
I am not able to rotate a M*N
matrix in anticlockwise direction. My code is working properly for 3*3
matrix, but when I try for any other case it is not working assume I am doing it for 4*4
matrix then only outer elements are rotating and inner 4 elements (i.e. 6,7,10,11) are not rotating. My input is 1-16 numbers as 4*4
matrix:
{ {1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16} }
static void antiRotateMatrix(int m, int n, int mat[][]) {
int row = 0, col = n - 1;
int prev, curr;
while (row < m && col < n) {
if (row + 1 == m || col - 1 == 0) {
break;
}
prev = mat[row + 1][col];
for (int i = col; i >= 0; i--) {
curr = mat[row][i];
mat[row][i] = prev;
prev = curr;
}
row++;
for (int i = row; i < m; i++) {
curr = mat[i][0];
mat[i][0] = prev;
prev = curr;
}
n--;
if (row < m) {
for (int i = n - 2; i <= col; i++) {
curr = mat[m - 1][i];
mat[m - 1][i] = prev;
prev = curr;
}
}
m--;
if (col <= n) {
for (int i = m - 1; i >= row; i--) {
curr = mat[i][col];
mat[i][col] = prev;
prev = curr;
}
}
col++;
}
for (int i = 0; i <= 3; i++) {
for (int j = 0; j <= 3; j++)
System.out.print(mat[i][j] + " ");
System.out.print("\n");
}
}
对于 4 * 4
,我应该得到
{{2,3,4,8},{1,7,11,12},{5,6,10,16},{9,13,14,15}}
但是我得到了
{{2,3,4,8},{1,6,7,12},{5,10,11,16},{9,13,14,15}}
推荐答案
在逆时针方向旋转每个图层 M * N
的矩阵元素:
Rotate matrix element of each layer M*N
in anticlockwise direction:
for z in range(r):#r = n times to rotate your element of each layer
top = 0
bottom = len(matrix)-1
left = 0
right = len(matrix[0])-1
while left < right and top < bottom: # anticlockwise rotation of each layer.
prev = matrix[top+1][right]
for i in range(right,left-1,-1):
curr = matrix[top][i]
matrix[top][i]= prev
prev =curr
top += 1
for i in range(top,bottom+1):
curr = matrix[i][left]
matrix[i][left]=prev
prev=curr
left += 1
for i in range(left,right+1):
curr =matrix[bottom][i]
matrix[bottom][i]=prev
prev = curr
bottom -=1
for i in range(bottom,top-1,-1):
curr = matrix[i][right]
matrix[i][right]=prev
prev = curr
right -=1
要顺时针转到> https://www.geeksforgeeks.org/rotate-matrix-元素/
我希望您可以将此 #python
代码应用于 #java
,逻辑是相同的.或在hackerrank中选择python.
I hope you can apply this #python
code to #java
, logic is the same. Or select python in hackerrank.
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