LeetCode Array Easy 189. Rotate Array

---恢复内容开始---

Description

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 = 0; i < k; i++)
            temp[i] = nums[length - k + i];
        for(int i = length - k -1; i >=0; i--)
            nums[i+k]=nums[i];
        for(int i = 0; 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, 0, nums.Length-1);
        Reverse(nums, 0, k-1);
        Reverse(nums, k, nums.Length-1);
    }
    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。