LeetCode 189. Rotate Array

分析

难度 易

来源

https://leetcode.com/problems/rotate-array/submissions/

题目

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?

解答

Runtime: 1 ms, faster than 52.53% of Java online submissions for Rotate Array.

package LeetCode;

public class L189_RotateArray {
    // Reverse the first n - k elements,
    // the last k elements, and then all the n elements.
    public void rotate(int[] nums, int k) {
        int len=nums.length;
        k=k%len;
        int temp=0;
        for(int i=0;i<(len-k)/2;i++){
            temp=nums[i];
            nums[i]=nums[len-k-1-i];
            nums[len-k-1-i]=temp;
        }
        for(int i=0;i<k/2;i++){
            temp=nums[len-k+i];//前半段下标0~len-k-1;后半段len-k~len-1
            nums[len-k+i]=nums[len-1-i];
            nums[len-1-i]=temp;
        }
        for(int i=0;i<len/2;i++){
            temp=nums[i];
            nums[i]=nums[len-1-i];
            nums[len-1-i]=temp;
        }
    }

    public static void main(String[] args){
        int[] nums={1,2,3,4,5,6,7};
        L189_RotateArray l189=new L189_RotateArray();
        l189.rotate(nums,3);
        for (int num:nums) {
            System.out.print(num+"\t");
        }
    }
}