[LeetCode]189.Rotate Array
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题目
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.
Hint:
Could you do it in-place with O(1) extra space?
思路一
跟字符串左右旋转操作一样的思路。
三次翻转法,第一次翻转前n-k个,第二次翻转后k个,第三次翻转全部。
代码一
/*-------------------------------------------- * 日期:2014-02-25 * 作者:SJF0115 * 题目: 189.Rotate Array * 网址:https://oj.leetcode.com/problems/rotate-array/ * 结果:AC * 来源:LeetCode * 博客: ------------------------------------------------*/ #include <iostream> #include <stdio.h> using namespace std; class Solution { public: void rotate(int nums[], int n, int k) { if(n <= 1){ return; }//if k = k % n; if(k <= 0){ return; }//if // 翻转前n-k个 Reverse(nums,0,n - k - 1); // 翻转后k个 Reverse(nums,n - k,n - 1); // 翻转全部 Reverse(nums,0,n - 1); } private: void Reverse(int nums[],int left,int right){ int tmp; for(int i = left,j = right;i < j;++i,--j){ // 交换 swap(nums[i],nums[j]); }//for } }; int main() { Solution solution; int A[] = {1,2,3,4,5,6,7}; int n = 7; int k = 2; solution.rotate(A,n,k); for(int i = 0;i < n;++i){ cout<<A[i]<<" "; }//for cout<<endl; return 0; }
思路二
充分利用Cycle Sort(圈排序)思路。
一个环总是要回到原来的位置。这与K和N的最小公倍数有关。 P步之后当PK== QN == LCM时环遍历完成。
- 我们可以很容易地看出,P <= N;
- 如果P是N的因子,这意味着该程序有N/P个环;
- 证明上面结论:p和q的最大公约数为1,p为q* n的因子,因此p是n的因子。
代码二
/*-------------------------------------------- * 日期:2014-02-25 * 作者:SJF0115 * 题目: 189.Rotate Array * 网址:https://oj.leetcode.com/problems/rotate-array/ * 结果:AC * 来源:LeetCode * 博客: ------------------------------------------------*/ #include <iostream> #include <stdio.h> using namespace std; class Solution { public: void rotate(int nums[], int n, int k) { if (n <= 0){ return; }//if k = k % n; if (k <= 0) { return; }//if int cycle = 0; int next = 0; int tmp = nums[next]; for ( int i = 0; i< n; i++) { next = (next + k) % n; swap(nums[next], tmp); // 如果一圈转完,向前移动一步 if (cycle == next) { next ++; cycle = next; tmp = nums[next]; }//if }//for } }; int main() { Solution solution; int A[] = {1,2,3,4,5,6,7}; int n = 7; int k = 2; solution.rotate(A,n,k); for(int i = 0;i < n;++i){ cout<<A[i]<<" "; }//for cout<<endl; return 0; }
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