189. Rotate Array

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 = 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]

Note:

• Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

 

My idea:pop insert 投机了

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        l=len(nums)
        k=k%l
        for i in range(k):
            a=nums.pop()
            nums.insert(0,a)

 

执行用时 : 184 ms, 在Rotate Array的Python3提交中击败了20.09% 的用户

内存消耗 : 12.9 MB, 在Rotate Array的Python3提交中击败了99.73% 的用户

 

学习一下别人的思路，四种方法很详细：

import java.util.Arrays;

class Solution {
    /**
     * 双重循环
     * 时间复杂度：O(kn)
     * 空间复杂度：O(1)
     */
    public void rotate_1(int[] nums, int k) {
        int n = nums.length;
        k %= n;
        for (int i = 0; i < k; i++) {
            int temp = nums[n - 1];
            for (int j = n - 1; j > 0; j--) {
                nums[j] = nums[j - 1];
            }
            nums[0] = temp;
        }
    }

    /**
     * 翻转
     * 时间复杂度：O(n)
     * 空间复杂度：O(1)
     */
    public void rotate_2(int[] nums, int k) {
        int n = nums.length;
        k %= n;
        reverse(nums, 0, n - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, n - 1);
    }


    private void reverse(int[] nums, int start, int end) {
        while (start < end) {
            int temp = nums[start];
            nums[start++] = nums[end];
            nums[end--] = temp;
        }
    }

    /**
     * 循环交换
     * 时间复杂度：O(n)
     * 空间复杂度：O(1)
     */
    public void rotate_3(int[] nums, int k) {
        int n = nums.length;
        k %= n;
        // 第一次交换完毕后，前 k 位数字位置正确，后 n-k 位数字中最后 k 位数字顺序错误，继续交换
        for (int start = 0; start < nums.length && k != 0; n -= k, start += k, k %= n) {
            for (int i = 0; i < k; i++) {
                swap(nums, start + i, nums.length - k + i);
            }
        }
    }

    /**
     * 递归交换
     * 时间复杂度：O(n)
     * 空间复杂度：O(n/k)
     */
    public void rotate(int[] nums, int k) {
        // 原理同上
        recursiveSwap(nums, k, 0, nums.length);
    }

    private void recursiveSwap(int[] nums, int k, int start, int length) {
        k %= length;
        if (k != 0) {
            for (int i = 0; i < k; i++) {
                swap(nums, start + i, nums.length - k + i);
            }
            recursiveSwap(nums, k, start + k, length - k);
        }
    }

    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}