LeetCode: 189. Rotate Array

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189. Rotate Array

LeetCode - Rotate Array

发现官网测试用例少了一个特殊情况：如果 k 大于两倍数组长度，提交的测试就会报错，但是官网却显示通过。

数组反转的方法也不错！反转时，直接使用界限的值进行加减比较简单省事！

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?

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/**
 * Runtime: 0 ms, faster than 100.00% of Java online submissions for Rotate Array.
 *
 * Memory Usage: 37.4 MB, less than 100.00% of Java online submissions for Rotate Array.
 *
 * Copy from: https://leetcode.com/problems/rotate-array/solution/[Rotate Array solution - LeetCode]
 */
public void rotate(int[] nums, int k) {
    if (Objects.isNull(nums) || nums.length == 0 || k % nums.length == 0) {
        return;
    }
    k %= nums.length;
    reverse(nums, 0, nums.length - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, nums.length - 1);
}

private void reverse(int[] nums, int start, int end) {
    while (start < end) {
        int temp = nums[start];
        nums[start] = nums[end];
        nums[end] = temp;
        start++;
        end--;
    }
}

/**
 * Runtime: 0 ms, faster than 100.00% of Java online submissions for Rotate Array.
 * <p>
 * Memory Usage: 38 MB, less than 85.58% of Java online submissions for Rotate Array.
 */
public void rotateExtraArray(int[] nums, int k) {
    if (Objects.isNull(nums) || nums.length == 0 ||  k % nums.length == 0) {
        return;
    }
    k %= nums.length;
    int[] temp = new int[k];
    for (int i = 0; i < k; i++) {
        temp[k - i - 1] = nums[nums.length - i - 1];
    }
    for (int i = nums.length - 1; i >= k; i--) {
        nums[i] = nums[i - k];
    }
    for (int i = 0; i < k; i++) {
        nums[i] = temp[i];
    }
}