184.Rotate Array
题目：
Given an array, rotate the array to the right by k steps, where k is non-negative.
给定一个数组，将数组向右旋转k步，其中k为非负数。
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.尝试尽可能多地提出解决方案，至少有3种不同的方法可以解决这个问题。
- Could you do it in-place with O(1) extra space?你能用O（1）额外空间就地做到吗？
解答：
方法一：用O（1）的额外空间，但是O（n2）的时间复杂度，太慢了
1 class Solution { 2 public void rotate(int[] nums, int k) { 3 for(int i=0;i<k;i++) 4 rotateOneStep(nums); 5 return; 6 } 7 8 private void rotateOneStep(int[] nums){ 9 int n=nums.length; 10 int lastNum=nums[n-1]; 11 for(int i=n-1;i>0;i--) 12 nums[i]=nums[i-1]; 13 nums[0]=lastNum; 14 } 15 }
方法二：O（n）空间复杂度——利用辅助数组
1 class Solution { 2 public void rotate(int[] nums, int k) { 3 int n=nums.length; 4 int[] tempNums=new int[n]; 5 for(int i=0;i<n;i++) 6 tempNums[i]=nums[i]; 7 for(int i=0;i<n;i++) 8 nums[(i+k)%n]=tempNums[i]; //(i+k)%n即为旋转后i元素的新位置 9 } 10 }
方法三：类似翻转字符串
1 class Solution { 2 public void rotate(int[] nums, int k) { 3 int n=nums.length; 4 if(nums.length==0 || (k%=n)==0) return; //k%=n使得k一定小于n，如果为0则不需要旋转 5 reverse(nums,0,n-k-1); 6 reverse(nums,n-k,n-1); 7 reverse(nums,0,n-1); 8 } 9 10 private void reverse(int[] nums,int start,int end){ 11 while(start<end){ 12 int temp=nums[start]; 13 nums[start]=nums[end]; 14 nums[end]=temp; 15 start++; 16 end--; 17 } 18 } 19 }
详解：
方法三分析：n=7，k=3
0 1 2 3 4 5 6 坐标
1 2 3 4 5 6 7 原数组
4 3 2 1 5 6 7 前n-k个数字翻转 [0,n-k-1]
4 3 2 1 7 6 5 后k个数字翻转 [n-k,n-1]
5 6 7 1 2 3 4 整个字符串翻转 [0,n-1]
注意k可能大于n，例如nums=[1,2]，k=3，旋转3位相当于旋转3%2=1位
