【LeetCode & 剑指offer刷题】数组题9：旋转数组（189. Rotate Array）

【LeetCode & 剑指offer刷题】数组题9：旋转数组（189. Rotate Array）

【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）

189. Rotate Array（相当于循环右移k位）

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?

相关题目： 循环左移

//方法一：使用额外的数组，索引i用i+k替换，然后把组织好的数组复制过去

// O(n), O(n)

/*class Solution

{

public:

    void rotate(vector& a, int k)

    {

        vector temp(a);

        int n = a.size();

        for(int i = 0; i

        {

            temp[(i+k)%n] = a[i]; //进行题意的rotated

        }

       

        a = temp; //将数据复制过去

    }

};*/

/*

方法二：多次反转法

Original List                   : 1 2 3 4 5 6 7

After reversing all numbers     : 7 6 5 4 3 2 1

After reversing first k numbers : 5 6 7 4 3 2 1

After revering last n-k numbers : 5 6 7 1 2 3 4 --> Result

分析：O(n) O(1)

*/

#include

class Solution

{

public :

    void rotate ( vector < int >& a , int k )

    {

        k %= a . size (); //以免k过大

       

        reverse ( a . begin (), a . end ()); //反转所有元素

        reverse ( a . begin (), a . begin ()+ k ); //反转前k个元素

        reverse ( a . begin ()+ k , a . end ()); //反转剩余元素（注意reverse反转区间为前闭后开，故这里为a.begin()+k）

    }

};

//方法三：使用循环替代，略

 

 

 

 

posted @ 2019-01-05 14:09 wikiwen 阅读( ...) 评论( ...) 编辑 收藏