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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.

 1 package LeetCode;
 2 
 3 public class L189_RotateArray {
 4     // Reverse the first n - k elements,
 5     // the last k elements, and then all the n elements.
 6     public void rotate(int[] nums, int k) {
 7         int len=nums.length;
 8         k=k%len;
 9         int temp=0;
10         for(int i=0;i<(len-k)/2;i++){
11             temp=nums[i];
12             nums[i]=nums[len-k-1-i];
13             nums[len-k-1-i]=temp;
14         }
15         for(int i=0;i<k/2;i++){
16             temp=nums[len-k+i];//前半段下标0~len-k-1;后半段len-k~len-1
17             nums[len-k+i]=nums[len-1-i];
18             nums[len-1-i]=temp;
19         }
20         for(int i=0;i<len/2;i++){
21             temp=nums[i];
22             nums[i]=nums[len-1-i];
23             nums[len-1-i]=temp;
24         }
25     }
26 
27     public static void main(String[] args){
28         int[] nums={1,2,3,4,5,6,7};
29         L189_RotateArray l189=new L189_RotateArray();
30         l189.rotate(nums,3);
31         for (int num:nums) {
32             System.out.print(num+"\t");
33         }
34     }
35 }

 

 

posted on 2018-11-18 22:36  flowingfog  阅读(82)  评论(0编辑  收藏