# 189. Rotate Array # 581. Shortest Unsorted Continuous Subarray

原创于 2018-10-27 16:50:26 发布 · 126 阅读

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本文深入探讨了数组旋转问题，提供了一种直接且易于理解的方法来实现数组的右旋操作。通过具体的实例，如输入数组[1,2,3,4,5,6,7]与k=3，展示如何逐步实现数组的旋转，并最终得到[5,6,7,1,2,3,4]。此外，还讨论了另一种特殊情况，当输入为[-1,-100,3,99]与k=2时，如何通过两次旋转得到[3,99,-1,-100]的结果。文章鼓励读者探索更多解决方案，至少提供了三种不同的解决策略。

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189. Rotate Array

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
还没看讨论区的答案，思路就是先将数组中最后一个数字保存在temp中，再通过循环将索引为0到size()-1的元素向右移，最后将temp的值赋给v[0].

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        if(k == 0)
            return;
        for(int i = 0; i<k; i++)
        {
            int temp = nums[nums.size()-1];
            for(int i = nums.size() -1; i>0; i--)
            {
                nums[i] = nums[i-1];
            }
            nums[0] = temp;
        }
    }
};