Title Description (medium difficulty)
A matrix rotated 90 degrees clockwise, and does not use extra space. Probably belong to find the law of the question, there is no general ideas, observe it.
A Solution
Can first transpose, then each column symmetrical about switching exchange.
public void rotate(int[][] matrix) {
//以对角线为轴交换
for (int i = 0; i < matrix.length; i ) {
for (int j = 0; j <=i; j ) {
if (i == j) {
continue;
}
int temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
//交换列
for (int i = 0, j = matrix.length - 1; i < matrix.length / 2; i , j--) {
for (int k = 0; k < matrix.length; k ) {
int temp = matrix[k][i];
matrix[k][i] = matrix[k][j];
matrix[k][j] = temp;
}
}
}
Time complexity: O (n²).
Complexity Space: O (1).
You can first axis transverse to the axis of symmetry of the line exchange, then switching to a diagonal.
Solution two
I put this link ideas posted over, inside comments also take the opportunity to have a chart posted over it and write well.
Circle around the loop interchange, it is wonderful!
public void rotate(int[][] matrix) {
int n=matrix.length;
for (int i=0; i<n/2; i )
for (int j=i; j<n-i-1; j ) {
int tmp=matrix[i][j];
matrix[i][j]=matrix[n-j-1][i];
matrix[n-j-1][i]=matrix[n-i-1][n-j-1];
matrix[n-i-1][n-j-1]=matrix[j][n-i-1];
matrix[j][n-i-1]=tmp;
}
}
Time complexity: O (n²).
Complexity Space: O (1).
total
This question is the subject of features were observed on it.
For more detailed explanations of popular leetcode.wang .
