力扣48
https://leetcode-cn.com/problems/rotate-image/
所谓原地旋转，就是不能新建一个数组
思路1：计算某个点距离上下左右的坐标，一次循环翻转4个值
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| |- - - - - - - - - - - - - - - - -|
| ↑ |
| toTop |
| ↓ |
| ← toLeft → o ← - - toRight - - →|
| ↑ |
| | |
| toBottom |
| | |
| ↓ |
|- - - - - - - - - - - - - - - - -|
旋转过后
|- - - - - - - - - - - - - - - - -|
| ↑ |
| toLeft |
| ↓ |
| ← - - toBottom - - → o ← toTop →|
| ↑ |
| | |
| toRight |
| | |
| ↓ |
|- - - - - - - - - - - - - - - - -|
四个一周期
点a旋转后占了b的位置，b占了c的位置，c占了d的位置，d一定会回到a的位置
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用个变量记录下来，或者python的话可以直接a, b, c, d = b, c, d, a
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| public void rotate(int[][] matrix) {
int row = (matrix.length % 2 == 0) ? (matrix.length / 2) : (matrix.length / 2 + 1);
for (int i = 0; i < row; i++) {
for (int j = 0; j < matrix[0].length / 2; j++) {
int toTop = i, toRight = matrix[0].length - j - 1, toBottom= matrix.length - i - 1, toLeft = j;
int temp = matrix[toTop][toLeft];
matrix[toTop][toLeft] = matrix[toRight][toTop];
matrix[toRight][toTop] = matrix[toBottom][toRight];
matrix[toBottom][toRight] = matrix[toLeft][toBottom];
matrix[toLeft][toBottom] = temp;
}
}
}
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思路2：转置+对称
顺时针：转置+左右对称
180°：上下对称+左右对称
逆时针：转置+上下对称
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| public void rotate2(int[][] matrix) {
for(int i=0;i<matrix.length;i++){
for(int j=0;j<i;j++){
int tem=matrix[j][i];
matrix[j][i]=matrix[i][j];
matrix[i][j]=tem;
}
}
for(int i=0;i<matrix.length;i++){
for(int j = 0; j < matrix.length / 2; j++){
int tem = matrix[i][matrix.length - 1 - j];
matrix[i][matrix.length - 1 - j] = matrix[i][j];
matrix[i][j] = tem;
}
}
}
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其他题目点击这里