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。
- 首先找到分割点
5
和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]
- 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