Recover Rotated Sorted Array
最后更新于：2022-04-02 01:08:14
# 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 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 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 &nums, vector::size_type start, vector::size_type end) {
for (vector::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 &nums) {
for (vector::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::size_type`替换为`int`
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