[LeetCode]189. Rotate Array

InfoSecPractitioner

于 2016-10-05 15:55:24 发布

阅读量269

点赞数

分类专栏： leetcode 文章标签： leetcode

leetcode 专栏收录该内容

64 篇文章

订阅专栏

本文介绍了如何将一个包含n个元素的数组向右旋转k步，并提供了三种不同的解决方案，包括一种空间复杂度为O(1)的方法。通过具体代码示例展示了不同方法的效率对比。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Easy

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.

space O(1):
2ms:

public void rotate(int[] nums, int k) {
        k %=  nums.length;
        for(int i=0;i<nums.length/2;i++){
            int tmp = nums[nums.length-1-i];
            nums[nums.length-1-i] = nums[i];
            nums[i] = tmp;
        }
        for(int j=0;j<k/2;j++){
            int tmp = nums[k-1-j];
            nums[k-1-j] = nums[j];
            nums[j] = tmp;
        }
        for(int w=k;w<(nums.length-k)/2+k;w++){
            int tmp = nums[k+nums.length-1-w];
            nums[k+nums.length-1-w] = nums[w];
            nums[w] = tmp;
        }
}

1ms:

public void rotate2(int[] nums, int k){
        k %= nums.length;
        int[] temp = new int[nums.length];
        System.arraycopy(nums, 0, temp, 0, nums.length-k);
        System.arraycopy(nums, nums.length-k, nums, 0, k);
        System.arraycopy(temp, 0, nums, k, nums.length-k);

}