leetcode 48. Rotate Image(矩阵旋转)
2017-04-20 00:46
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You are given an n x n 2D matrix representing an image.Rotate the image by 90 degrees (clockwise).Follow up:
Could you do this in-place?将矩阵顺时针旋转90度。思路:先在纸上画一个3x3的矩阵,在画出它旋转90度之后的图形,观察其变化------------------------------->,本质上是坐标的变化,毕竟到时候操作的是个二维数组
(0,0)(0,1)(0,2) (2,0)(1,0)(0,0)(1,0)(1,1)(1,2) (2,1)(1,1)(0,1)(2,0)(2,1)(2,2) ----------------->(2,2)(1,2)(0,2),可以看出,行的下标只变过一次就不变了,而列的下标是依次增加,可以看出,从下往上把每一行单独拿出来,从左往右放到每一列上去,就能得到顺时针旋转90度的图形了
参考代码如下,代码只是可以ac而已,没有做优化一开始先初始化二维数组a,并为每一列分配空间,因为该方法给定的返回值是void,所以最后再把旋转得到的矩阵a重新赋值给matrixpublic void rotate(int[][] matrix) {
int[][] a = new int[matrix.length][];
for(int i = 0; i<matrix.length; i++)
{
a[i] = new int[matrix[i].length];
}
int row = matrix.length - 1;
for(int i = 0; i < matrix.length; i++)
{
for(int j = 0; j < matrix[i].length; j++)
{
a[j][i] = matrix[row][j];
}
row--;
}
for(int i = 0; i < matrix.length; i++)
{
for(int j = 0; j < matrix[i].length; j++)
{
matrix[i][j] = a[i][j];
}
}
}
Could you do this in-place?将矩阵顺时针旋转90度。思路:先在纸上画一个3x3的矩阵,在画出它旋转90度之后的图形,观察其变化------------------------------->,本质上是坐标的变化,毕竟到时候操作的是个二维数组
(0,0)(0,1)(0,2) (2,0)(1,0)(0,0)(1,0)(1,1)(1,2) (2,1)(1,1)(0,1)(2,0)(2,1)(2,2) ----------------->(2,2)(1,2)(0,2),可以看出,行的下标只变过一次就不变了,而列的下标是依次增加,可以看出,从下往上把每一行单独拿出来,从左往右放到每一列上去,就能得到顺时针旋转90度的图形了
参考代码如下,代码只是可以ac而已,没有做优化一开始先初始化二维数组a,并为每一列分配空间,因为该方法给定的返回值是void,所以最后再把旋转得到的矩阵a重新赋值给matrixpublic void rotate(int[][] matrix) {
int[][] a = new int[matrix.length][];
for(int i = 0; i<matrix.length; i++)
{
a[i] = new int[matrix[i].length];
}
int row = matrix.length - 1;
for(int i = 0; i < matrix.length; i++)
{
for(int j = 0; j < matrix[i].length; j++)
{
a[j][i] = matrix[row][j];
}
row--;
}
for(int i = 0; i < matrix.length; i++)
{
for(int j = 0; j < matrix[i].length; j++)
{
matrix[i][j] = a[i][j];
}
}
}
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