题目
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?
我的尝试
没有更好的思路，除了暴力循环，直接看思路，思路很不错，也容易理解，自己写一下流程：
1 反转整个数组，此时shift的k个元素就到了前面
2. 反转前面k个元素，这个就是反转后k个的最终结果
3. 前面的元素是平移过去，在第一步反转后，再次反转，达到平移的效果。
class Solution {
public void rotate(int[] nums, int 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[] nums,int start,int end){
while (start<end){
int tmp=nums[start];
nums[start]=nums[end];
nums[end]=tmp;
start++;
end--;
}
}
}