【LeetCode】189. Rotate Array 解題報告

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189. Rotate Array / Medium

Given an array, rotate the array to the right by k steps, where k is non-negative.

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 <= 10^5
  • -2^31 <= nums[i] <= 2^31 - 1
  • 0 <= k <= 10^5

Solution 1: Insert & Erase

思路

既然題目要求要 O(1) 的空間複雜度，也就是要 In-Place，那就只好使用 vector 操作了。

要進行旋轉，其實就是把後面 k 個搬到前面，但 vector 在最前面 insert 效能花費比較大（雖說怎麼也避不開就是了…），所以我就將前面 n（vector size）-k 個元素搬到後面，再 erase 前面的 n-k 個元素，就達成旋轉的效果了。

效能

Complexity

  • Time Complexity: O(N^2), vector::erase takes O(N^2)
  • Space Complexity: O(1)

LeetCode Result

Code

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class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        if(k >= nums.size()) k = k % nums.size();
        
        int s = nums.size()-k;
        nums.insert(nums.end(), nums.begin(), nums.begin()+s);
        nums.erase(nums.begin(), nums.begin()+s);
    }
};

Solution 2: std::reverse

思路

將前面 n-k 個元素旋轉，再將後面 k 個旋轉，最後整個 vector 旋轉，就是答案了。

效能

Complexity

  • Time Complexity: O(N), where vector::reverse takes O(N)
  • Space Complexity: O(1)

LeetCode Result

Code

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class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        if(k >= nums.size()) k = k % nums.size();
        reverse(nums.begin(), nums.begin() + nums.size() - k); 
        reverse(nums.begin() + nums.size() - k, nums.end());
        reverse(nums.begin(), nums.end());
    }
};

Solution 3: std::rotate

思路

std::rotate 可以直接達成題目需要的功能。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(1)

LeetCode Result

Code

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class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        if(k >= nums.size()) k = k % nums.size();
        std::rotate(nums.begin(), nums.begin()+nums.size()-k, nums.end());
    }
};