voidrotate(vector<int>& nums,intk) {intn =nums.size();if(n ==0|| k ==0)return;//求最大公约数inta = n, c = k, b = a %c;while(b !=0) { a=c; c=b; b= a %c; }//依次替换while(c--) {intoldNum =nums[c];intcurPos = (c + k) %n;while(curPos !=c) {intcu...
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]. 就是把后k个数字进行旋转,放到前面去。 如果使用辅助空间的话就会非常简单: 1、再开一个空间大小相同的数组,然后先放后面的k个元素,再放前面的len-k个。 2、开一个大小为Math.min(k, len - ...
public void rotate(int[] nums, int k) { if (k > nums.length) { k = k % nums.length; } int tmp[] = (int[]) nums.clone(); for (int i = 0; i < k; i++) { nums[i] = tmp[tmp.length - k + i]; } for (int i = 0; i < (tmp.length - k); i++) { nums[k...
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之Rotate Array 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]. Note: Try to come up as many solutions as you can, there are at least 3 different ...
LeetCode 189: Rotate Array (Java) 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]. 两种解法: 第一种是原地swap,记录下被挤掉的数再接着swap,需要一些时间举栗子...
[Array]Rotate Array Problem link: https://leetcode.com/explore/interview/card/top-interview-questions-easy/92/array/646/leetcode.com/explore/interview/card/top-interview-questions-easy/92/array/646/ Solutions: Given an array, rotate the array to the right byksteps, wherekis non-negative....
length; k = k % len; int count = 0; // 记录交换位置的次数,n个同学一共需要换n次 for(int start = 0; count < len; start++) { int cur = start; // 从0位置开始换位子 int pre = nums[cur]; do{ int next = (cur + k) % len; int temp = nums[next]; // 来到角落... ...
Given an array, rotate the array to the right by k steps, where k is non-negative. Example 1: Input:[1,2,3,4,5,6,7]and k=3Output:[5,6,7,1,2,3,4]Explanation:rotate1steps to theright:[7,1,2,3,4,5,6]rotate2steps to theright:[6,7,1,2,3,4,5]rotate3steps to theri...
以二分搜索为基本思路简要来说: nums[0] <= nums[mid](0 - mid不包含旋转)且nums[0] <= target <= nums[mid] 时 high 向前规约; nums[mid] < nums[0](0 - mid包含旋转),target <= nums[mid] < nums[ C++ 二分查找 598 238.2K 221Sweetie...