文章目录
题目描述
思路
思路1:利用辅助数组
思路2:原地翻转数组
复杂度
思路1:
时间复杂度:
空间复杂度:
思路2:
时间复杂度:
空间复杂度:
Code
class Solution {
public:
/**
*
* @param nums To be operated array
* @param k Specifies the number of rotations
*/
void rotate(vector<int>& nums, int k) {
int n = nums.size();
vector<int> temp(n);
for (int i = 0; i < n; ++i) {
temp[(i + k) % n] = nums[i];
}
for (int i = 0; i < n; ++i) {
nums[i] = temp[i];
}
}
};
class Solution {
public:
/**
*
* @param nums To be operated array
* @param k Specifies the number of rotations
*/
void rotate(vector<int> &nums, int k) {
int n = nums.size();
reverse(nums, 0, n - 1);
reverse(nums,0, k % n - 1);
reverse(nums, k % n, n - 1);
}
void reverse(vector<int>& nums, int left, int right) {
int n = nums.size();
while (left < right) {
swap(nums[left], nums[right]);
left++;
right--;
}
}
};