189. Rotate Array

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?

// Using extra space
void rotate(vector<int>& nums, int k) { // time: O(n); space: O(n)
    vector<int> tmp = nums;
    int n = nums.size();
    for (int i = 0; i < n; ++i) {
        nums[(i + k) % n] = tmp[i];
    }
}

// Constant version of the previous method
void rotate(vector<int>& nums, int k) { // time: O(n); space: O(1)
    if (nums.empty() || (k %= nums.size()) == 0) return;
    int start_idx = 0, idx = 0, pre = 0, cur = nums[0], n = nums.size();
    for (int i = 0; i < n; ++i) {
        pre = cur;
        idx = (idx + k) % n;
        cur = nums[idx];
        nums[idx] = pre;
        if (idx == start_idx) {
            start_idx = ++idx;
            cur = nums[start_idx];
        }
    }
}

// Reverse method
void rotate(vector<int>& nums, int k) { // time: O(n); space: O(1)
    if (nums.empty() || ((k %= nums.size()) == 0)) return;
    int n = nums.size();
    reverse(nums.begin(), nums.end());
    reverse(nums.begin(), nums.begin() + k);
    reverse(nums.begin() + k, nums.end());
}

// Reverse method
void rotate(vector<int>& nums, int k) { // time: O(n); space: O(1)
    if (nums.empty() || ((k %= nums.size()) == 0)) return;
    int n = nums.size();
    reverse(nums.begin(), nums.begin() + n - k);
    reverse(nums.begin() + n - k, nums.end());
    reverse(nums.begin(), nums.end());
}

// Swap numbers to correct positions and keep doing so for the next part of array
void rotate(vector<int>& nums, int k) { // time: O(n); space: O(1)
    int n = nums.size(), start = 0;
    while (n && (k %= n)) {
        for (int i = 0; i < k; ++i) {
            swap(nums[i + start], nums[n - k + i + start]);
        }
        n -= k;
        start += k;
    }
}

61. Rotate List