記錄下自己拙計的算法之旅 LeetCode Rotate Array

記錄下自己拙計的算法之旅

**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.

語言 ruby

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

空間複雜度為O(n)，時間複雜度O(n)，拙計~~

另一種方法

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

但是時間複雜度為O(n^2)，空間為O(1)，直接Time Limit Exceeded，更拙計 = =