问题

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?

解题思路

直接根据编程珠玑一书中得思路而来的，双手旋转法，具体如下：

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public void rotate(int[] nums, int k) {
    if (null == nums || nums.length <= 0 || k <= 0) {
        return;
    }
    k = k % nums.length;
    reverse(nums, 0, nums.length - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, nums.length - 1);
}

private void reverse(int[] nums, int start, int end) {
    while (start < end) {
        int tmp = nums[start];
        nums[start] = nums[end];
        nums[end] = tmp;
        ++start;
        --end;
    }
}

分析

  • 时间复杂度为O(n)，空间复杂度为O(1);