（Java）leetcode-189 Rotate Array

topic

【旋转数组】
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?

Thinking

k greater than the length of the array, the array after each cycle nums.length reduction, it should be effective k k% nums.length.
Well, this time the problem is transformed into - "k-bit array after moving to the forefront of the array."
The method can be done by reversing the array:
[1,2,3,4,5,6,7]
After the first k bits, and the former 7-k-bit flip-situ:
[4,3,2,1] + [ 7,6,5]
At this time, the entire array and then inverted once:
[5,6,7,1,2,3,4]
time complexity: O (n)
complexity of space: O (1)

Code

class Solution {
    public void rotate(int[] nums, int k) {
        
        k = k%nums.length;//k比数组长度大的情况，每nums.length次循环后数组还原，所以有效的k应该是k%nums.length
              
        reverse(nums, nums.length-k, nums.length-1);
        reverse(nums, 0, nums.length-k-1);
        reverse(nums, 0, nums.length-1);
    }

    public void reverse(int[] nums, int start, int end){//用来反转数组中指定的一段
    	for( int temp; start < end ; start++,end--){   		
    		temp = nums[end];
    		nums[end] = nums[start];
    		nums[start] = temp;
    	}
    }
}

Present the results

Runtime: 0 ms, faster than 100.00% of Java online submissions for Rotate Array.
Memory Usage: 38.2 MB, less than 30.38% of Java online submissions for Rotate Array.