Leet Code OJ 189. Rotate Array [Difficulty: Easy]
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题目:
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.
Hint:
Could you do it in-place with O(1) extra space?
翻译:
旋转一个n个元素的数组右移k步。
例如,n是7,k是3,数组是 [1,2,3,4,5,6,7],旋转后变成 [5,6,7,1,2,3,4]。
提示:尝试尽可能的使用多种方案,这里至少有3种不同的方式来解决这个问题。
你可以在原数组上完成操作,并且使用O(1)的额外空间吗?
分析:
首先要理解题目所谓的旋转。[1,2,3,4,5,6,7]旋转右移1步,将变成[7,1,2,3,4,5,6]。由于移动元素的过程中,会发生覆盖,以下方案使用了一个临时数组来备份原数组,再进行移动。O(1)的空间复杂度的方案,笔者暂时没想出来,欢迎大家留言交流下自己的实现方案。
代码:
public class Solution { public void rotate(int[] nums, int k) { if(k>nums.length){ k=k%nums.length; } int[] nums2=new int[nums.length]; for(int i=0;i<nums.length;i++){ nums2[i]=nums[i]; } for(int i=0;i<nums.length;i++){ int m=i-k; if(m<0){ m+=nums.length; } nums[i]=nums2[m]; } }}
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