LeetCode 189. Rotate Array 题解

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Total Accepted: 76162 Total Submissions: 352104 Difficulty: 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.

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public class Solution {
    public void rotate(int[] nums, int k) {
        //method one

        int n = nums.length;
        int[] tempNums = new int[n];
        for(int l = 0; l < n; l++){
            tempNums[l] = nums[l];
        }
         // 根据k和n的大小进行判断
        if ( k > n) {//递归求解
            
            k = k % n;
           rotate(nums, k);
            
        } else if (k < n) {
        
            
            if(k != 0) {
    
                for ( int i = 0; i < k; i++){
                    nums[i] = tempNums[n - k + i];
                }
                
                for ( int j = 0; j < n - k ; j++){
                    nums[k + j] = tempNums[j];
                }
                
            }
       
        }
        
        //method two insert
        int n = nums.length;
        int temp = 0;
       
        // 根据k和n的大小进行判断
        if ( k > n) {//递归求解
            
           k = k % n;
           rotate(nums, k);
            
        } else if (k < n){
            
            int[] tempNums = new int[n - k];
            for(int l = 0; l < n-k; l++){
                tempNums[l] = nums[l];
            }
            
            if(k != 0) {
                // 从k开始前插到数组
                for ( int i = 0; i < k; i++){//控制插入个数
                    
                    temp = nums[n - k + i];
                    
                    // for (int j  = n - k; j > 0; j-- ){//控制移动个数
                    //      nums[j + i ] = nums[j + i - 1];
                    // }
                    
                    nums[i] = temp;
                  
                }
                
                for ( int j = 0; j < n - k ; j++){
                    nums[k + j] = tempNums[j];
                }
                
            }
        }
        
        //method three time limit
        
        int n = nums.length;
        int temp = 0;
       
        // 根据k和n的大小进行判断
        if ( k > n) {//递归求解
            
           k = k % n;
           rotate(nums, k);
            
        } else if (k < n){

            if(k != 0) {
                // 从k开始前插到数组
                for ( int i = 0; i < k; i++){//控制插入个数
                    
                    temp = nums[n - k + i];
                    
                    for (int j  = n - k; j > 0; j-- ){//控制移动个数
                         nums[j + i ] = nums[j + i - 1];
                    }
                    
                    nums[i] = temp;
                  
                }
                
            
        }

}