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.

[show hint]

Hint:
Could you do it in-place with O(1) extra space?

Related problem: Reverse Words in a String II

题意：

从给定的第K个位置翻转数组。

思路一：

将数组复制一份，之后将复制的数组从第K个位置一次放回到原数组中。

代码：

class Solution {public:    void rotate(vector<int>& nums, int k) {                int n = nums.size();        vector<int> result(n);                if(n==0 || k<=0){            return;        }                for(int i=0; i<n; i++){            result[i] = nums[i];        }                for(int i=0; i<n; i++){            nums[(i+k)%n] = result[i];   //(i+k)%n相当于产生数组“环”        }    }};

思路二：

一个数字一个数字的交换。

代码：

class Solution {public:    void rotate(vector<int>& nums, int k) {        int n = nums.size();                if(n==0 || k<=0){            return;        }           int cur = 0;        int tmp = 0;        int numRotated = nums[0];        int count = 0;        int start = 0;                while(count<n){            do{                tmp = nums[(cur+k)%n];                nums[(cur+k)%n] = numRotated;                numRotated = tmp;                cur = (cur+k)%n;                count++;            }while(cur!=start);                        start++;            cur = start;            numRotated = nums[cur];        }    }};

思路三：

先将前n-k个元素翻转，再讲n-k到n个元素翻转，之后将全部n个元素翻转

4 3 2 1    7 6 5 =》 5 6 7 1 2 3 4

代码：

class Solution {public:    void rotate(vector<int>& nums, int k) {        int n = nums.size();        k = k%n;                reverse(nums.begin(), nums.begin()+n-k);                reverse(nums.begin()+n-k, nums.end());                reverse(nums.begin(), nums.end());    }};

public class Solution {    public void rotate(int[] nums, int k) {        int n = nums.length;        k %= n;                reverse(nums, 0, n-k-1);                reverse(nums, n-k, n-1);                reverse(nums, 0, n-1);    }        public void reverse(int[] nums, int start, int end){        while(start<end){            int tmp = nums[start];            nums[start] = nums[end];            nums[end] = tmp;            start++;            end--;        }    }}

思路四：

先将最后k个元素与前k个元素交换，之后前半部分有序，处理后半部分即可

5 6 7   4 1 2 3 

class Solution {public:    void rotate(int nums[], int n, int k)     {        for (; k = k%n; n -= k, nums += k)        {            // Swap the last k elements with the first k elements.             // The last k elements will be in the correct positions            // but we need to rotate the remaining (n - k) elements             // to the right by k steps.            for (int i = 0; i < k; i++)            {                swap(nums[i], nums[n - k + i]);            }        }    }};

思路五：

同样分成两个子串，之后对部分处理。

1 5 6 7   2 3 4

class Solution {public:    void rotate(int nums[], int n, int k)     {        if ((n == 0) || (k <= 0) || (k%n == 0))        {            return;        }        k = k%n;        // Rotation to the right by k steps is equivalent to swapping         // the two subarrays nums[0,...,n - k - 1] and nums[n - k,...,n - 1].        int start = 0;        int tmp = 0;        while (k > 0)        {            if (n - k >= k)            {                // The left subarray with size n - k is longer than                 // the right subarray with size k. Exchange                 // nums[n - 2*k,...,n - k - 1] with nums[n - k,...,n - 1].                for (int i = 0; i < k; i++)                {                    tmp = nums[start + n - 2*k + i];                    nums[start + n - 2*k + i] = nums[start + n - k + i];                    nums[start + n - k + i] = tmp;                }                // nums[n - 2*k,...,n - k - 1] are in their correct positions now.                // Need to rotate the elements of nums[0,...,n - k - 1] to the right                 // by k%n steps.                n = n - k;                k = k%n;            }            else            {                // The left subarray with size n - k is shorter than                 // the right subarray with size k. Exchange                 // nums[0,...,n - k - 1] with nums[n - k,...,2*(n - k) - 1].                for (int i = 0; i < n - k; i++)                {                    tmp = nums[start + i];                    nums[start + i] = nums[start + n - k + i];                    nums[start + n - k + i] = tmp;                }                // nums[n - k,...,2*(n - k) - 1] are in their correct positions now.                // Need to rotate the elements of nums[n - k,...,n - 1] to the right                 // by k - (n - k) steps.                tmp = n - k;                n = k;                k -= tmp;                start += tmp;            }        }    }};

转载链接：https://leetcode.com/discuss/27387/summary-of-c-solutions