Given an array, rotate the array to the right by k steps, where k is non-negative.
Follow up:
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?
Example 1:
Input: nums = [1,2,3,4,5,6,7], 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: nums = [-1,-100,3,99], 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]
Constraints:
- 1 <= nums.length <= 2 * 104
- -231 <= nums[i] <= 31 - 1
- 0 <= k <= 105
/**
* @param {number[]} nums
* @param {number} k
* @return {void} Do not return anything, modify nums in-place instead.
*/
var rotate = function(nums, k) {
k %= nums.length;
while (k > 0) {
nums.unshift(nums.pop());
k--;
}
};
var rotate = function(nums, k) {
k = k % nums.length;
reverse(nums, 0, nums.length - 1);
reverse(nums, 0, k - 1);
reverse(nums, k, nums.length - 1);
};
var reverse = (nums, start, end) => {
while (start < end) {
let tmp = nums[start];
nums[start] = nums[end];
nums[end] = tmp;
start++;
end--;
}
};
說明舉例
Original List : 1 2 3 4 5 6 7
After reversing all numbers : 7 6 5 4 3 2 1
After reversing first k numbers : 5 6 7 4 3 2 1
After revering last n-k numbers (Result) : 5 6 7 1 2 3 4
k%=nums.length 並分三次 reverse 即可得到所要的
Complexity Analysis
- Time complexity: O(n)
- Space complexity: O(1)