README.md

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?

Solutions (Rust)

1. Brute Force

impl Solution {
    pub fn rotate(nums: &mut Vec<i32>, k: i32) {
        for _ in 0..k {
            for i in (1..nums.len()).rev() {
                nums.swap(i, i - 1);
            }
        }
    }
}

2. Store Last k Numbers

impl Solution {
    pub fn rotate(nums: &mut Vec<i32>, k: i32) {
        let k = k as usize % nums.len();
        let last_k = nums[(nums.len() - k)..].to_vec();
        for i in (k..nums.len()).rev() {
            nums.swap(i, i - k);
        }
        nums.splice(..k, last_k);
    }
}

3. Reverse

impl Solution {
    pub fn rotate(nums: &mut Vec<i32>, k: i32) {
        let k = k as usize % nums.len();
        nums.reverse();
        nums[..k].reverse();
        nums[k..].reverse();
    }
}

4. Cyclic Replacements

impl Solution {
    pub fn rotate(nums: &mut Vec<i32>, k: i32) {
        let k = k as usize % nums.len();
        let mut start = 0;
        let mut cnt = 0;
        while cnt < nums.len() {
            let mut cur = start;
            let mut prev = nums[cur];
            loop {
                let next = (cur + k) % nums.len();
                let temp = nums[next];
                nums[next] = prev;
                cur = next;
                prev = temp;
                cnt += 1;
                
                if start == cur {
                    break;
                }
            }
            start += 1;
        }
    }
}