解决:可以计算a1位置移动后的下一个位置a2 = (a1 + k) % n,然后让nums[a2] = nums[a1],依次类推,直到下一个位置an回到起点位置 也就是 a,这样循环下去,总会将所有元素替换掉吧。 3.我试着编写出一个循环程序,提交后发现程序存在严重漏洞,如果 n % k = 0 或者 k % n = 0,我们执行这个程序会发...
LeetCode笔记:189. Rotate Array 大意: 旋转一个有n个元素的数组,右移k步。 比如,n = 7,k = 3,数组 [1,2,3,4,5,6,7] 就被旋转为 [5,6,7,1,2,3,4]。 思路: 旋转本身没太多特别好说的,我的做法是用另一个数组来记录旋转后的内容,然后复制回原数组。当然记录时是从第nums.length-k个元素...
题目链接: Rotate Array : leetcode.com/problems/r 轮转数组: leetcode-cn.com/problem LeetCode 日更第 86 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满 发布于 2022-04-14 09:29 力扣(LeetCode) Python 算法 赞同添加评论 分享喜欢收藏申请转载 ...
LeetCode Rotate Array 翻转数组 题意:给定一个数组,将该数组的后k位移动到前n-k位之前。(本题在编程珠玑中第二章有讲) 思路: 方法一:将后K位用vector容器装起来,再移动前n-k位到后面,再将容器内k位插到前面。 Rotate Array 方法二:将后k位自身旋转,再将前n-k位自身旋转,在将整个数组旋转。例如:[1...
LeetCode之“数组”:Rotate Array 题目链接 题目要求: Rotate an array ofnelements to the right byksteps. 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 ...
LeetCode Rotate Array Rotate Array Total Accepted: 12759 Total Submissions: 73112 My Submissions Question Solution 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]....
void rotationArray(int* arr, int k, int n) { int temp[k]; // 临时数组 int i,j; // 1. 保存数组 arr 中的前 k 个元素到临时数组 temp 中 for( i = 0;i < k;i++) { temp[i] = arr[i]; } // 2. 将数组中的其余元素向前移动k个位置 ...
leetcode-189-Rotate Array Explanation:rotaterotate7,1, Example 2: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 Input:[-1,-100,3,99]and k=2Output:[3,99,-1,-100]Explanation:rotate1steps to the right:[99,-1,-100,3]rotate2steps to the right:[3,99,-1,-100]...
Rotate the cell array by 270 degrees. B = rot90(A,3) B =3×3×2 cell arrayB(:,:,1) = {'g'} {'d'} {'a'} {'h'} {'e'} {'b'} {'i'} {'f'} {'c'} B(:,:,2) = {'p'} {'m'} {'j'} {'q'} {'n'} {'k'} {'r'} {'o'} {'l'} ...
You swap the smallest array linearly towards its proper location, since the blocks behind it are in the proper location you can forget about them. What remains of the larger array is now the smallest array, which you rotate in a similar manner, until the smallest side shrinks to 0 elements...