[leetcode] 189. Rotate Array
Easy
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?
题目大意:将数组中的数字向右循环移动k位,比如数组{1,2,3,4,5,6,7},循环右移k=3位变成{5,6,7,1,2,3,4}
方法:
使用数字交换的方法实现数字右移。先把整个数组反转,然后把数组的前k位和后n-k位分别反转。
要注意如果k的值比数组长度大,那么就只要右移多出的那部分就可以了,也就是移动k%n位。
代码如下:
class Solution { public: void rotate(vector<int>& nums, int k) { int n = nums.size(); k = k % n; if (k == 0 || nums.size() <= 1)return; for (int i = 0; i < (n-1) / 2 + 1; ++i) { swap(nums[i], nums[n - 1 - i]); } for (int j = 0; j < (k-1) / 2 + 1; ++j) { swap(nums[j], nums[k - 1 - j]); } for (int j = k; j < (n + k - 1) / 2 + 1; ++j) { swap(nums[j], nums[n - 1 - j + k]); } return; } };
数组反转的地方也可以直接用vector的reverse函数,代码如下:
class Solution { public: void rotate(vector<int>& nums, int k) { int n = nums.size(); k = k % n; if (k == 0 || nums.size() <= 1)return; reverse(nums.begin(),nums.end()); reverse(nums.begin(),nums.begin()+k); reverse(nums.begin()+k,nums.end()); return; } };