Recover Rotated Sorted Array

Question

  1. Given a rotated sorted array, recover it to sorted array in-place.
  3. Example
  4. [4, 5, 1, 2, 3] -> [1, 2, 3, 4, 5]
  6. Challenge
  7. In-place, O(1) extra space and O(n) time.
  9. Clarification
  10. What is rotated array:
  12.     - For example, the orginal array is [1,2,3,4], The rotated array of it can be [1,2,3,4], [2,3,4,1], [3,4,1,2], [4,1,2,3]

首先可以想到逐步移位，但是这种方法显然太浪费时间，不可取。下面介绍利器『三步翻转法』，以[4, 5, 1, 2, 3]为例。

  1. 首先找到分割点5和1
  2. 翻转前半部分4, 5为5, 4，后半部分1, 2, 3翻转为3, 2, 1。整个数组目前变为[5, 4, 3, 2, 1]
  3. 最后整体翻转即可得[1, 2, 3, 4, 5]

由以上3个步骤可知其核心为『翻转』的in-place实现。使用两个指针，一个指头，一个指尾，使用for循环移位交换即可。

Java

  1. public class Solution {
  2.     /**
  3.      * @param nums: The rotated sorted array
  4.      * @return: The recovered sorted array
  5.      */
  6.     public void recoverRotatedSortedArray(ArrayList<Integer> nums) {
  7.         if (nums == null || nums.size() <= 1) {
  8.             return;
  9.         }
  11.         int pos = 1;
  12.         while (pos < nums.size()) { // find the break point
  13.             if (nums.get(pos - 1) > nums.get(pos)) {
  14.                 break;
  15.             }
  16.             pos++;
  17.         }
  18.         myRotate(nums, 0, pos - 1);
  19.         myRotate(nums, pos, nums.size() - 1);
  20.         myRotate(nums, 0, nums.size() - 1);
  21.     }
  23.     private void myRotate(ArrayList<Integer> nums, int left, int right) { // in-place rotate
  24.         while (left < right) {
  25.             int temp = nums.get(left);
  26.             nums.set(left, nums.get(right));
  27.             nums.set(right, temp);
  28.             left++;
  29.             right--;
  30.         }
  31.     }
  32. }

C++

  1. /**
  2.  * forked from
  3.  * http://www.jiuzhang.com/solutions/recover-rotated-sorted-array/
  4.  */
  5. class Solution {
  6. private:
  7.     void reverse(vector<int> &nums, vector<int>::size_type start, vector<int>::size_type end) {
  8.         for (vector<int>::size_type i = start, j = end; i < j; ++i, --j) {
  9.             int temp = nums[i];
  10.             nums[i] = nums[j];
  11.             nums[j] = temp;
  12.         }
  13.     }
  15. public:
  16.     void recoverRotatedSortedArray(vector<int> &nums) {
  17.         for (vector<int>::size_type index = 0; index != nums.size() - 1; ++index) {
  18.             if (nums[index] > nums[index + 1]) {
  19.                 reverse(nums, 0, index);
  20.                 reverse(nums, index + 1, nums.size() - 1);
  21.                 reverse(nums, 0, nums.size() - 1);
  23.                 return;
  24.             }
  25.         }
  26.     }
  27. };

源码分析

首先找到分割点，随后分三步调用翻转函数。简单起见可将vector<int>::size_type替换为int