Question: http://leetcode.com/2010/04/rotating-array-in-place.html Rotate a one-dimensional array of n elements to the right by k steps. For instance, with n=7 and k=3, the array {a, b, c, d, e, f, g} is rotated to {e, f, g, a, b, c, d}...
Related problem:Reverse Words in a String II 解法1:一个最简单的想法就是每次将数组后移一步,一共循环k次。满足in-place,但是时间复杂度为O(k*n)。 classSolution {public:voidrotate(vector<int>& nums,intk) {intn =nums.size();if(n <2|| k <1)return;for(inti =1; i <= k; i++) {...
Question: http://leetcode.com/2010/04/rotating-array-in-place.html Rotate a one-dimensional array of n elements to the right by k steps. For instance, with n=7 and k=3, the array {a, b, c, d, e, f, g} is rotated to {e, f, g, a, b, c, d} // O (n)publicvoidro...
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. [show hint] Hint: Could you do it in-place with O(1)...
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?
class Solution: def rotate(self, nums: List[int], k: int) -> None: """ Do not return anything, modify nums in-place instead. """ n: int = len(nums) # 计算最后有多少数字会被移动到数组开始 k %= n # 翻转整个数组 Solution.reverse(nums, 0, n - 1) # 翻转前 k 个数字 Soluti...
For example, withn= 7 andk= 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?
Hint: Could you do it in-place with O(1 LeetCode编程练习 - Rotate Array学习心得 题目: Rotate an array ofn elements to the right byk steps. For example, withn = 7 andk = 3, the array[1,2,3,4,5,6,7] is rotated to[5,6,7,1,2,3,4]. 将n个元素的数组按k级旋转。例如,n...
Could you do it in-place with O(1) extra space? 要完成的函数: void rotate(vector<int>& nums, int k) 说明: 1、这道题给定一个vector,要求将这个vector最右边的元素调到最左边,重复这个动作k次,最终结果仍然存放在nums中。要求空间复杂度为O(1)。
class Solution(object): def rotate(self, nums, k): """ :type nums: List[int] :type k: int :rtype: void Do not return anything, modify nums in-place instead. """ length = len(nums) k = k % length for i in range(length / 2): nums[i],nums[length - i - 1] = nums[len...