# [LeetCode] 189. Rotate Array

## Description

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

• 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:

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:

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

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

i -> (i + k) % nums.length

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

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


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

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);
}

posted @ 2020-05-27 19:05  arcsinW  阅读(106)  评论(0编辑  收藏  举报