LeetCode 189——旋转数组，用O(1)的空间复杂度实现

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;
}