LeetCode Array Easy 189. Rotate Array

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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]
Example 2:
Input: [-1,-100,3,99] and 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]
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?

问题描述：给定一个长度为n的数组和一个非负整数k，向右旋转k部。

思路，首先这里k是可以大于数组的长度的。我采用最暴力的解法，这种解法效率很低，提交之后只能到13%-85% 效率不固定

public void Rotate(int[] nums, int k) {

        k = k % nums.Length;

        int[] temp = new int[k];

        if(k== nums.Length)

            return;

        int length = nums.Length;

        for(int i = ; i < k; i++)

            temp[i] = nums[length - k + i];

        for(int i = length - k -; i >=; i--)

            nums[i+k]=nums[i];

        for(int i = ; i < k; i++)

            nums[i]=temp[i];

    }

解法二采用反转的方式，先把所有的数组反转，然后反转前k个元素，再反转后n-k个元素

public class Solution {

    public void Rotate(int[] nums, int k) {

        k = k % nums.Length;

        Reverse(nums, , nums.Length-);

        Reverse(nums, , k-);

        Reverse(nums, k, nums.Length-);

    }

    private void Reverse(int[] arr, int start, int end){

        while(start < end){

            int temp = arr[start];

            arr[start]=arr[end];

            arr[end]=temp;

            start++;

            end--;

        }

    }

}

但是实测发现，后一种方法效率还不如第一种方法。

巴特西

LeetCode Array Easy 189. Rotate Array

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