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
Navigator_Z2 小时前
LeetCode //C - 1209. Remove All Adjacent Duplicates in String II
c语言·算法·leetcode
土司大王4 小时前
LeetCode hot100——合并两个有序链表
算法·leetcode·链表
wabs66611 小时前
关于栈【力扣1047. 删除字符串中的所有相邻重复项的思考】
数据结构·c++·算法·leetcode··代码随想录
evans在进步14 小时前
LeetCode 53 最大子数组和:一次遍历掌握 Kadane 算法
算法·leetcode·职场和发展
Nil20814 小时前
leetcode 230二叉搜索树中第k小的元素
算法·leetcode·职场和发展
旖旎夜光14 小时前
LeetCode 69:x 的平方根(二分查找) —— 题解
数据结构·c++·算法·leetcode·二分查找
学习星球15 小时前
【LeetCode算法题精讲】图算法精讲——从图遍历到拓扑排序
数据结构·算法·leetcode·图搜索
Nil2081 天前
leetcode 98验证二叉搜索树
算法·leetcode·职场和发展
时针滴滴答啊1 天前
最大子数组和
算法·leetcode·职场和发展
重生之后端学习1 天前
15. 三数之和[中等]✅
java·数据结构·算法·leetcode·职场和发展