记录下自己拙计的算法之旅 LeetCode Rotate Array

记录下自己拙计的算法之旅

**LeetCode
Rotate Array：**
Rotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

语言 ruby

def rotate(nums, k)    num_cp = nums.clone    pos = k % nums.length    start_pos = 0    nums.length.times do        nums[pos] = num_cp[start_pos]        start_pos += 1        pos += 1        pos %= nums.length     endend

空间复杂度为O(n)，时间复杂度O(n)，拙计~~

另一种方法

def rotate2 nums, k    nums_length = nums.length    k = k % nums_length    return if k == 0 || nums_length == 1    k.times do        temp = nums[nums_length-1]        point = nums_length-1        point.times do            nums[point] = nums[point-1]            point -= 1        end        nums[0] = temp    endend

但是时间复杂度为O(n^2)，空间为O(1)，直接Time Limit Exceeded，更拙计 = =