Leetcode:Rotate Array

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于 2015-03-05 17:12:12 发布

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分类专栏： algorithm 文章标签： algorithm leetcode

本文链接：https://blog.youkuaiyun.com/xiaomutiao/article/details/44085189

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本文介绍了一种在O(1)额外空间复杂度下旋转数组的方法。通过两次遍历，首先整体反转整个数组，然后分别反转前k个元素和后n-k个元素，实现数组的高效旋转。

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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.

思路：两次遍历，第一次先整体rotate，第二次再前k个rotate，后n-k个rotate

时间复杂度O(n), 空间复杂度O(1)

代码：

class Solution {
public:
    void rotate(int nums[], int n, int k) {
        if(nums == NULL || n < 1 || k < 1)return;
        k = k % n;
        for(int i = 0, j = n - 1; i < j; ++i, --j){
            swap(nums[i], nums[j]);
        }
        for(int i = 0, j = k - 1; i < j; ++i, --j){
            swap(nums[i], nums[j]);
        }
        for(int i = k, j = n - 1; i < j; ++i, --j){
            swap(nums[i], nums[j]);
        }
    }
};