leetcode - 189. Rotate Array

Description

Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.

Example 1:

复制代码
Input: nums = [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

复制代码
Input: nums = [-1,-100,3,99], k = 2
Output: [3,99,-1,-100]
Explanation: 
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Constraints:

复制代码
1 <= nums.length <= 10^5
-2^31 <= nums[i] <= 2^31 - 1
0 <= k <= 10^5

Follow up:

复制代码
Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
Could you do it in-place with O(1) extra space?

Solution

Solved after help...

For a space o ( 1 ) o(1) o(1) solution, one straightforward way is: start with 0, use current value to set the new index, and then update the current value with the original value in the new index, and keep going until we replaced all the numbers.

This works for most cases, but doesn't work when k = k % n, for example, when n=4 and k=2, if we start with 0, then we jump to 2, then we jump back to 0. This could create a loop.

To avoid this loop, before starting, we should keep track of where we started, and once we found we arrived at start again, we increase start by 1 and keep going.

Time complexity: o ( n ) o(n) o(n)

Space complexity: o ( 1 ) o(1) o(1)

Code

python3 复制代码
class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        k %= len(nums)
        rotate_cnt = 0
        start = 0
        while rotate_cnt < len(nums):
            new_index, value_to_set = start, nums[start]
            while True:
                new_index = (new_index + k) % len(nums)
                ori_value = nums[new_index]
                nums[new_index] = value_to_set
                value_to_set = ori_value
                rotate_cnt += 1
                if new_index == start:
                    break
            start += 1
相关推荐
YuTaoShao1 小时前
【LeetCode 每日一题】1653. 使字符串平衡的最少删除次数——(解法一)前后缀分解
算法·leetcode·职场和发展
VT.馒头1 小时前
【力扣】2727. 判断对象是否为空
javascript·数据结构·算法·leetcode·职场和发展
老鼠只爱大米3 小时前
LeetCode经典算法面试题 #46:全排列(回溯、交换、剪枝等五种实现方案详细解析)
算法·leetcode·剪枝·回溯·全排列·stj算法
im_AMBER4 小时前
Leetcode 114 链表中的下一个更大节点 | 删除排序链表中的重复元素 II
算法·leetcode
历程里程碑4 小时前
普通数组----轮转数组
java·数据结构·c++·算法·spring·leetcode·eclipse
pp起床4 小时前
贪心算法 | part02
算法·leetcode·贪心算法
sin_hielo4 小时前
leetcode 1653
数据结构·算法·leetcode
YuTaoShao4 小时前
【LeetCode 每日一题】3634. 使数组平衡的最少移除数目——(解法二)排序 + 二分查找
数据结构·算法·leetcode
Q741_1475 小时前
C++ 优先级队列 大小堆 模拟 力扣 703. 数据流中的第 K 大元素 每日一题
c++·算法·leetcode·优先级队列·
木井巳5 小时前
【递归算法】二叉搜索树中第K小的元素
java·算法·leetcode·深度优先·剪枝