[LeetCode]189.Rotate Array
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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