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