LeetCode 189——旋转数组，用O(1)的空间复杂度实现

最新推荐文章于 2025-05-27 16:45:46 发布

饿二饿二饿

最新推荐文章于 2025-05-27 16:45:46 发布

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本文介绍了一种基于数组长度和旋转位数最大公约数的数组旋转算法。通过具体实例展示了如何实现数组元素的有效移动，以及算法背后的数学原理。适用于理解数组操作及算法优化。

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旋转的次数是，数组长度n 和旋转位数 k的最大公约数

折腾了俩小时才琢磨出来，笨死我了

#include<iostream>
#include<vector>
#include<string>
#include<ctime>
#include<queue>
#include<cassert>
#include<algorithm>

using namespace std;


class Solution {
public:
	void rotate(vector<int>& nums, int k) {
		int cur_index,next_index,cur;
		int n = nums.size();
		cout <<"n:"<< n << endl;
		cout << "k:" << k<< endl;
		//i属于[0.......n-k-1]时，应该往后移k位，+k
		//i属于[n-k......n-1]时，下标变为i-（n-k）
		k = k%n;
		if (n == 0 || n == 1||k==0)
			return;

		//旋转的次数就是n和k的最大公约数
		//求n和k的
		int a = n,b = k;
		while (a != b)
		{
			if (a>b)
			{
				a = a - b;
			}
			else
			{
				b = b - a;
			}
		}

		int rotate_count = a;
		cout << "rotate:" << rotate_count << endl;

		for (int i = 1; i <= rotate_count; i++)
		{
			cur_index = n - k - i;
			cur = nums[cur_index];
			next_index = cur_index + k;
			int temp = nums[next_index];
			nums[next_index] = cur;
			cur = temp;
			cur_index = next_index;

			while (cur_index != n - k - i)
			{
				if (cur_index >= 0 && cur_index <= n - k - 1)
				{
					next_index = cur_index + k;
					int temp = nums[next_index];
					nums[next_index] = cur;
					cur = temp;
					cur_index = next_index;

				}
				if (cur_index >= n - k && cur_index <= n - 1)
				{
					next_index = cur_index - (n - k);
					int temp = nums[next_index];
					nums[next_index] = cur;
					cur = temp;
					cur_index = next_index;

				}
			}


			cout << "开始第"<<i<<"部分旋转" << endl;
			for (int i = 0; i < n; i++)
				cout << nums[i] << " ";
			cout << endl;
		}
	}
};
int main()
{
	vector<int> arr = { 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54 };
	//vector<int> arr = { 1,2,3,4,5,6,7,8,9,10 };
	Solution().rotate(arr,45);
	return 0;
}