【LeetCode】Rotate Array 旋转数组

题目

Rotate Array

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.

题目大意

旋转数组，给定数组长度n，旋转位移k，将数组元素循环右移k位。

思路

三步翻转法，《编程珠玑》一书上有讲，步骤如下：

1. 将数组从k位置分为前后两半；
2. 翻转前半部分；
3. 翻转后半部分；
4. 翻转整个数组；

图示如下：

解答

Java代码如下，注意遍历时候的边界不要出错：

    public void rotate(int[] nums, int k) {       if(k==0 || nums==null ) return;        k = k%nums.length;        int tmp;        for (int i = 0; i < (nums.length-k)/2; i++) {            tmp = nums[i];            nums[i] = nums[nums.length-k-1-i];            nums[nums.length-k-1-i] = tmp;        }        /*for (int i : nums) {            System.out.print(i+"-");        }        System.out.println();*/        for (int i = 0; i < k/2; i++) {            tmp = nums[i+nums.length-k];            nums[i+nums.length-k] = nums[nums.length-1-i];            nums[nums.length-1-i] = tmp;        }        /*for (int i : nums) {            System.out.print(i+" ");        }        System.out.println();*/        for (int i = 0; i < nums.length/2; i++) {            tmp = nums[i];            nums[i] = nums[nums.length-1-i];            nums[nums.length-1-i] = tmp;        }    }