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.

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Special thanks to  for adding this problem and creating all test cases.

 

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题解：

数组移动问题。可以参考<编程珠玑>里利用三次reverse达成数组向左移动的方法。向右移动k step相当于向左移动数组长度nums.length - k。 要注意提前取摸，使k成为一个合理的值。

Time Complexity - O(n)， Space Complexity - O(1)。

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

 

Update:

依然是三步反转法，先反转整个数组，再翻转0 ~ k - 1， 最后翻转 k ~ n - 1

public class Solution {    public void rotate(int[] nums, int k) {        if(nums == null || nums.length == 0)            return;        int n = nums.length;        k %= n;                reverse(nums, 0, n - 1);        reverse(nums, 0, k - 1);        reverse(nums, k, n - 1);    }        private void reverse(int[] nums, int lo, int hi) {        while(lo < hi)            swap(nums, lo++, hi--);    }        private void swap(int[] nums, int lo, int hi) {        int tmp = nums[lo];        nums[lo] = nums[hi];        nums[hi] = tmp;    }}

 

二刷: 

三步反转法.

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

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

 

三刷:

Java:

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

 

 

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