189. Rotate Array【easy】

189. Rotate Array【easy】

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

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

 解法一：

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int len = nums.size();
        
        if (len == 0 || k % len == 0)
        {
            return;
        }
        
        int mid = len - k % len;
        
        reverseArray(nums, 0, mid - 1);
        reverseArray(nums, mid, len - 1);
        reverseArray(nums, 0, len - 1);
    }
    
    void reverseArray(vector<int>& nums, int start, int end)
    {
        int s = start;
        int e = end;
        
        while (s <= e)
        {
            int temp = nums[s];
            nums[s] = nums[e];
            nums[e] = temp;
            s++;
            e--;
        }
    }
};

注意：

1、边界值检查，避免对0取余和长度不合法

2、先分别翻转，再总体翻转，注意下标

解法二：

class Solution 
    {
    public:
        void rotate(int nums[], int n, int k) 
        {
            k = k%n;
    
            // Reverse the first n - k numbers.
            // Index i (0 <= i < n - k) becomes n - k - i.
            reverse(nums, nums + n - k);
            
            // Reverse tha last k numbers.
            // Index n - k + i (0 <= i < k) becomes n - i.
            reverse(nums + n - k, nums + n);
            
            // Reverse all the numbers.
            // Index i (0 <= i < n - k) becomes n - (n - k - i) = i + k.
            // Index n - k + i (0 <= i < k) becomes n - (n - i) = i.
            reverse(nums, nums + n);
        }
    };

解法三：

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

先整体搞，再分开搞