Given an array, rotate the array to the right by k steps, where k is non-negative.
Example：
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]
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?
解释下题目：
数组的数字往后推移k位，如果超出了最后则接到头部
1. 倒置的方法
实际耗时：0ms
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[] nums, int start, int end) {
while (start < end) {
int tmp = nums[start];
nums[start] = nums[end];
nums[end] = tmp;
start++;
end--;
}
}
踩过的坑：[-1] k=2
我自己想到的思路是有三种，第一种是，搞个函数，然后这个函数的作用是每次给数组后退1格，然后如果是k的话就重复k次，这样时间是O(nk) 不需要额外的空间。第二种是再开辟一个同样大小的数组，很简单。最后一种就是上面代码里的，其实这种rotate的题目最后都可以用倒序来实现。
时间复杂度O(n)
空间复杂度O(1)
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