[array] leetcode - 48. Rotate Image - Medium
leetcode - 48. Rotate Image - Medium
descrition
You are given an n x n 2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).
Note:
You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.
Example 1 :
Given input matrix =
[
[1,2,3],
[4,5,6],
[7,8,9]
],
rotate the input matrix in-place such that it becomes:
[
[7,4,1],
[8,5,2],
[9,6,3]
]
Example 2:
Given input matrix =
[
[ 5, 1, 9,11],
[ 2, 4, 8,10],
[13, 3, 6, 7],
[15,14,12,16]
],
rotate the input matrix in-place such that it becomes:
[
[15,13, 2, 5],
[14, 3, 4, 1],
[12, 6, 8, 9],
[16, 7,10,11]
]
解析
参见代码。小技巧:矩阵的对角线可以唯一确定一个矩阵。
code
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution{
public:
void rotate(vector<vector<int> >& matrix){
//rotateNonInplace(matrix);
rotateNonInplace(matrix);
}
/*
(si,sj) ** (si,sj+k) *** (si,ej)
* *
* *
* (si+k,ej)
(ei-k,sj) *
* *
* *
(ei,sj)*** (ei,ej-k) ** (ei,ej)
*/
// time-O(n^2), space-O(1)
void rotateInplace(vector<vector<int> >& matrix){
int n = matrix.size();
int si = 0, sj = 0; // the top-left corner
int ei = n-1, ej = n-1; // the down-right corner
// (si, sj), (ei, ej)
while(si <= ei && sj<=ej){
for(int k=0; sj+k<=ej; k++){
int temp = matrix[si][sj+k];
matrix[si][sj+k] = matrix[ei-k][sj];
matrix[ei-k][sj] = matrix[ei][ej-k];
matrix[ei][ej-k] = matrix[si+k][ej];
matrix[si+k][ej] = temp;
}
si++;
sj++;
ei--;
ej--;
}
}
// time-O(n^2), space-O(n^2)
void rotateNonInplace(vector<vector<int> >& matrix){
int n = matrix.size();
vector<vector<int> > assit(n, vector<int>(n, 0));
// put the i-row in matrix to (n-1-i)-column in assit
for(int i=0; i<n; i++){
for(int k=0; k<n; k++){
assit[k][n-1-i] = matrix[i][k];
}
}
// copy assit to matrix
for(int i=0; i<n; i++){
for(int j=0; j<n; j++){
matrix[i][j] = assit[i][j];
}
}
}
};
int main()
{
return 0;
}
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