189. Rotate Array.md

189. Rotate Array

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?

Solution

最简单的和次简单都一次过，一个是耗时不耗内存；一个是不耗时耗内存，见code1.

code2是时间O(n)，空间O(1)的方法。

Code1

class Solution {
    
    public void rotate2(int[] nums, int k) {
        
        int len = nums.length;
        int tmp = 0;
        if(len<=1)
            return;
        
        k = k % len;
        for(;k>0;k--)
        {
            tmp = nums[len-1];
            for(int i=len-1;i>0;i--)
                nums[i] = nums[i-1];
            nums[0] = tmp;
        }
        return;
        
    }
    
    public void rotate(int[] nums, int k) {
        
        int len = nums.length;
        
        if(len<=1)
            return;
        
        k = k % len;
        int[] tmp = new int[k];
        for(int i=len-k;i<len;i++)
        {
            tmp[i-len+k] = nums[i];
        }
        
        for(int i=len-k-1;i>=0;i--)
        {
            nums[i+k] = nums[i];
        }
        
        for(int i=0;i<k;i++)
        {
            nums[i] = tmp[i];
        }
        return;
    }
}

Code2

class Solution {
    private void swap(int[] nums, int i, int j)
    {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
    
    private void reverse(int st, int ed, int[] nums)
    {
        while(st<ed)
        {
            swap(nums,st,ed);
            st ++ ;
            ed -- ;
        }
    }
    
    public void rotate(int[] nums, int k) {
        
        int len = nums.length;
        k = k % len;
        if(k==0 || len <= 1)
            return;
        
        reverse(0,len-1,nums);
        reverse(0,k-1,nums);
        reverse(k,len-1,nums);
        return ;
    }
}