LeetCode.189 Rotate Array

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于 2018-09-05 09:29:53 发布

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本文链接：https://blog.youkuaiyun.com/yige__cxy/article/details/82414449

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本文介绍了一种高效的数组旋转算法，该算法能在O(1)额外空间内实现数组元素的右移k位操作。通过实例演示了如何使用反转方法来达到目的，并提供了C++代码实现。

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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

代码：（效率超级高）

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int size = nums.size();
        if(size <= 1) return;
        k = k % size;
        reverse(nums.begin() + size - k, nums.end());
        reverse(nums.begin(), nums.begin() + size - k);
        reverse(nums.begin(), nums.end());
    }
};