# Recover Rotated Sorted Array

## Question

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]


1. 首先找到分割点51
2. 翻转前半部分4, 55, 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]

### 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;
}
}
}
};