leetcode------Rotate Array

标题: Rotate Array
通过率: 18.4%
难度: 简单

Rotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

[show hint]

Hint:
Could you do it in-place with O(1) extra space?

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

第一种思路,k等于多少就循环多少次:意思是若k=3则所有元素向后移动三次,越界的则进行放在第一位操作,

具体看代码:

 1 public class Solution {
 2     public void rotate(int[] nums, int k) {
 3         int length=nums.length;
 4         int value=0;
 5         for(int i=0;i<k;i++){
 6             value=nums[length-1];
 7             for(int j=nums.length-2;j>=0;j--){
 8                 nums[j+1]=nums[j];
 9             }
10             nums[0]=value;
11         }
12     }
13 }

上述思路明显不是高效的,下面对数据进行分析,

1.2.3.4.5

当k=3时结果是3,4,5,1,2

把结果逆序下 可以看出来是   2,1,5,4,3逆序结果与正序进行对比

1,2,3,4,5

2,1,5,4,3

发现在k-1的位置前后都是逆序的,

第二种分析。

原数据逆序

5,4,3,2,1

3,4,5,1,2

也是在k-1位置前后进行逆序,

那么我只用对原数据做如下操作就可得到结果:

1、原数据逆序

2,k-1之前逆序

3,k-1之后逆序

直接看代码:

 1 public class Solution {
 2     public void rotate(int[] nums, int k) {
 3         int length=nums.length;
 4         k%=length;
 5         res(nums,0,length-1);
 6         res(nums,0,k-1);
 7         res(nums,k,length-1);
 8         
 9     }
10     public void res(int []arraynum,int start,int end){
11         while(start<=end){
12             int temp=arraynum[start];
13             arraynum[start]=arraynum[end];
14             arraynum[end]=temp;
15             start++;
16             end--;
17         }
18     }
19 }

 

posted @ 2015-03-03 16:33  pku_smile  阅读(180)  评论(0编辑  收藏  举报