Leetcode OJ 189. Rotate Array

Rotate an array of n elements to the right by K steps.

For example, with n = 7 and k = 3, the array is [1,2,3,4,5,6,7] rotated to [5,6,7,1,2,3,4] .

Note:
Try to come up as many solutions as can, there is at least 3 different ways to solve this problem.

[Show hint]

Hint:
Could 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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"Idea One"

This topic means moving n = (k%nums.length) elements at the tail of the array to the front of the array. One of the simplest ideas is to move the next n elements forward in turn. The code is as follows:

1  Public classSolution {2      Public voidRotateint[] Nums,intk) {3         if(nums==NULL|| Nums.length <= 0)return;4         intn = k%nums.length;5         6          for(inti = 0; I < n; i++){7              for(intj = nums.length-n + i; J > i; j--){8                 inttemp = Nums[j-1];9NUMS[J-1] =Nums[j];TenNUMS[J] =temp; One             } A         } -     } -}

The complexity of the space is O (1), but the complexity of the time is high, how to improve it?

"Thinking Two"

For example: [1,2,3,4,5,6,7] k = 3

The array that first flips 5 of the data before it is [4,3,2,1,5,6,7]

The resulting array of 5 and later data is [4,3,2,1,7,6,5]

The entire array is then flipped to get [5,6,7,1,2,3,4]

The code is as follows:

1  Public classSolution {2      Public voidRotateint[] Nums,intk) {3         if(nums==NULL|| Nums.length <= 0 | | k%nums.length==0)return;4         intLength =nums.length;5K = k%length;6         7Reversal (nums, 0, length-k-1);8Reversal (Nums, length-k, length-1);9Reversal (nums, 0, Length-1);Ten     } One      Public voidReversal (int[] Nums,intIintj) { A         intt = 0; -          while(I < J && I >= 0){ -t =Nums[i]; theNums[i] =Nums[j]; -NUMS[J] =T; -i++; -j--; +              -         } +     } A}

Leetcode OJ 189. Rotate Array