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]
.
Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
Hint:
Could you do it in-place with O(1) extra space?
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.
新题抢先刷,这道题标为Easy,应该不是很难,题目明确要求了空间复杂度为O(1),那么肯定不能建立新的数组,所以只能覆盖原数组的值,仔细观察题目中给的那个例子,k = 3说是移动了三步,怎么移的呢,原来是每次把最末尾的数移动到开头,移三次就得到最终结果。代码如下:
class Solution { public: void rotate(int nums[], int n, int k) { for (int i = 0; i < k; ++i) { moveRight(nums, n); } } void moveRight(int nums[], int n) { if (n == 0) return; int last = nums[n - 1]; for (int i = n - 1; i >= 1; --i) nums[i] = nums[i - 1]; nums[0] = last; } };
题目说至少有三种解法,暂时只想到一种,容我慢慢想想~~
时间: 2024-08-05 09:58:11