题目:
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?
链接:https://leetcode.com/problems/rotate-image/
分析:这里我提供两种解法
法一:需要额外的空间复杂度O(MN),时间复杂度为O(MN),旋转90度就是将[i,j]放在新数组的[j, n-1-i]位置处
int n = matrix.size(); vector<vector<int> > tmp = matrix; for(int i = 0; i < n; i++){ for(int j = 0; j < n; j++){ matrix[j][n-i-1] = tmp[i][j]; } }
法二:将一副图像旋转90度就是将其上下对折,然后再按照对角线对折,It is amazing!!时间复杂度仍然为O(MN),但不需要额外的空间了。
class Solution { public: void rotate(vector<vector<int> > &matrix) { int n = matrix.size(); /* vector<vector<int> > tmp = matrix; for(int i = 0; i < n; i++){ for(int j = 0; j < n; j++){ matrix[j][n-i-1] = tmp[i][j]; } }*/ for(int i = 0; i < n; i++){ // 先上下对折 for(int j = 0; j < n/2; j++) swap(matrix[i][j], matrix[i][n-1-j]); } for(int i = 0; i < n; i++){ // 后按照对角线对折就能顺时针旋转90度了 for(int j = 0; j < n-1-i; j++) swap(matrix[i][j], matrix[n-1-j][n-i-1]); } } };
题目:
Rotate an array of n elements to the right by k steps.
For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7]
is
rotated to [5,6,7,1,2,3,4]
.
链接:https://leetcode.com/problems/rotate-array/
分析:这里仍提供两种方法,法一需要额外的空间O(N),法二利用对折就可以了,方法是先对折前面n-k个元素,然后对折后面k个元素,最后对折这n个元素就可以了,it is amazing!!
class Solution { public: void rotate(int nums[], int n, int k) { /* k = k%n; int *a = new int[k]; // 需要空间复杂度O(k) for(int i = 0; i < k; i++) a[i] = nums[n-k+i]; for(int i = n-k-1; i >= 0; i--) nums[i+k] = nums[i]; for(int i = 0; i < k; i++) nums[i] = a[i]; delete []a;*/ k = k % n; for(int i = 0; i < (n-k)/2; i++) // 先对折前面的n-k个元素 swap(nums[i], nums[n-k-1-i]); for(int i = 0; i < k/2; i++) swap(nums[n-k+i], nums[n-1-i]); // 对折后面的k个元素 for(int i = 0; i < n/2; i++) swap(nums[i], nums[n-1-i]); // 最后对折这n个元素 } };
时间: 2024-11-05 15:51:44