# Recover Rotated Sorted Array ### Source - lintcode: [(39) Recover Rotated Sorted Array](http://lintcode.com/en/problem/recover-rotated-sorted-array/) ~~~ Given a rotated sorted array, recover it to sorted array in-place. Example [4, 5, 1, 2, 3] -> [1, 2, 3, 4, 5] Challenge In-place, O(1) extra space and O(n) time. Clarification What is rotated array: - 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` 1. 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]` 1. 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 ~~~ public class Solution { /** * @param nums: The rotated sorted array * @return: The recovered sorted array */ public void recoverRotatedSortedArray(ArrayList<Integer> nums) { if (nums == null || nums.size() <= 1) { return; } int pos = 1; while (pos < nums.size()) { // find the break point if (nums.get(pos - 1) > nums.get(pos)) { break; } pos++; } myRotate(nums, 0, pos - 1); myRotate(nums, pos, nums.size() - 1); myRotate(nums, 0, nums.size() - 1); } private void myRotate(ArrayList<Integer> nums, int left, int right) { // in-place rotate while (left < right) { int temp = nums.get(left); nums.set(left, nums.get(right)); nums.set(right, temp); left++; right--; } } } ~~~ ### C++ ~~~ /** * forked from * http://www.jiuzhang.com/solutions/recover-rotated-sorted-array/ */ class Solution { private: void reverse(vector<int> &nums, vector<int>::size_type start, vector<int>::size_type end) { for (vector<int>::size_type i = start, j = end; i < j; ++i, --j) { int temp = nums[i]; nums[i] = nums[j]; nums[j] = temp; } } public: void recoverRotatedSortedArray(vector<int> &nums) { for (vector<int>::size_type index = 0; index != nums.size() - 1; ++index) { if (nums[index] > nums[index + 1]) { reverse(nums, 0, index); reverse(nums, index + 1, nums.size() - 1); reverse(nums, 0, nums.size() - 1); return; } } } }; ~~~ ### 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`