【LeetCode】16.Array and String — Rotate Array 旋转数组

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

Example 1:

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]

Example 2:

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?
  
  题目的解释中已经给的很清楚，在这里提供三种解法，其中第一和第二种算法Accept，第三种由于占用空间过多无法通过，写一下提供参考。
  法一：用栈的思维，先进后出，后出前进，调用C++自带函数，可以很简单的实现算法，该方法Accept。
  

   1 void rotate(vector<int>&nums, int k)
   2 {
   3     int size = nums.size();
   4     int record;
   5     int counter = k;
   6     while (counter>0)
   7     {
   8         record = nums[size - 1];
   9         nums.pop_back();
  10         nums.insert(nums.begin(), record);
  11         counter--;
  12     }
  13 }

  
  法二：利用取余规律进行数组数据按要求重排，方法简单，Accept。

  1 void rotate(vector<int>&nums, int k)
  2 {
  3     if (nums.size() == 0 || k <= 0)return;
  4     vector<int> space(nums);
  5     for (int i = 0; i<nums.size(); i++)
  6         nums[(i + k) % nums.size()] = space[i];
  7 }

  法三:严格按照Explanation中的步骤一步步执行，消耗空间多，无法Accept。

   1 void rotate(vector<int>nums, int k) //时间复杂度较高,LeetCode无法通过
   2 {
   3     int size = nums.size();
   4     int record;
   5     int counter = k;
   6     while (counter>0)
   7     {
   8         record = nums[size-1];
   9         for (int j = size - 1; j > 0; --j)
  10         {
  11             nums[j] = nums[j-1];
  12         }
  13         nums[0] = record;
  14         counter--;
  15     }
  16 }

   

转载于:https://www.cnblogs.com/hu-19941213/p/11011029.html