Recover Rotated Sorted Array

Question: (39) 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
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。

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