leetcode-189 Rotate Array

189. Rotate Array

Description

Given an array, rotate the array to the right by k steps, where k is non-negative.

Follow up:

  • 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?

Example 1:

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Input: nums = [1,2,3,4,5,6,7], 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:

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Input: nums = [-1,-100,3,99], 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]

Constraints:

  • 1 <= nums.length <= 2 * 10^4
  • It’s guaranteed that nums[i] fits in a 32 bit-signed integer.
  • k >= 0

Analyse

将数组nums的后k位翻转到数组起始位置

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Input:  [1,2,3,4,5,6,7], k = 3
Output: [5,6,7,1,2,3,4]  // 5, 6, 7翻转到数组首部

申请临时空间

原数组与翻转后的数组下标有对应关系
i -> (i + k) % nums.length

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public void rotate(int[] nums, int k) {
    int len = nums.length;
    k = k % len;

    int[] tmp = new int[len];

    for (int i = 0; i < len; ++i) {
        tmp[(i + k) % len] = nums[i];
    }

    for (int j = 0; j < len; ++j) {
        nums[j] = tmp[j];
    }
}

reverse三次

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k = 3
1 2 3 4 5 6 7   // expected 5 6 7 1 2 3 4
4 3 2 1 5 6 7   // reverse 1 2 3 4
4 3 2 1 7 6 5   // reverse 5 6 7
5 6 7 1 2 3 4   // reverse 全部

另一种三次reverse

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k = 3
1 2 3 4 5 6 7   // expected 5 6 7 1 2 3 4
4 3 2 1 5 6 7   // reverse 1 2 3 4
7 6 5 1 2 3 4   // reverse 全部
5 6 7 1 2 3 4   // reverse 5 6 7
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public void reverse(int[] nums, int low, int high) {
    while (low < high) {
        int tmp = nums[low];
        nums[low] = nums[high];
        nums[high] = tmp;
        ++low;
        --high;
    }
}

public void rotate(int[] nums, int k) {
    k %= nums.length;
    reverse(nums, 0, nums.length - k - 1);
    reverse(nums, nums.length - k, nums.length-1);
    reverse(nums, 0, nums.length-1);
}