题目描述
题目链接:969. Pancake Sorting
Given an array of integers A, We need to sort the array performing a series of pancake flips.
In one pancake flip we do the following steps:
- Choose an integer k where 0 <= k < A.length.
- Reverse the sub-array A[0…k]. For example, if A = [3,2,1,4] and we performed a pancake flip choosing k = 2, we reverse the sub-array [3,2,1], so A = [1,2,3,4] after the pancake flip at k = 2.
Return an array of the k-values of the pancake flips that should be performed in order to sort A. Any valid answer that sorts the array within 10 * A.length flips will be judged as correct.
例子
例子 1
Input: A = [3,2,4,1] Output: [4,2,4,3] Explanation: We perform 4 pancake flips, with k values 4, 2, 4, and 3. Starting state: A = [3, 2, 4, 1] After 1st flip (k = 4): A = [1, 4, 2, 3] After 2nd flip (k = 2): A = [4, 1, 2, 3] After 3rd flip (k = 4): A = [3, 2, 1, 4] After 4th flip (k = 3): A = [1, 2, 3, 4], which is sorted. Notice that we return an array of the chosen k values of the pancake flips.
例子 2
Input: A = [1,2,3] Output: [] Explanation: The input is already sorted, so there is no need to flip anything. Note that other answers, such as [3, 3], would also be accepted.
Note
1 <= A.length <= 100
1 <= A[i] <= A.length
- All integers in
A
are unique (i.e.A
is a permutation of the integers from1
toA.length
).
解题思路
这道题涉及到 pancakeSort()
, 一次 flip 表示反转前 k 个元素。要重复这个操作直至数组排好序。通过这个方法,我们可以想到,每一次先找到当前没排好序的最后一个位置的元素,假设在 i 处,即: A[i] = target
,先将那个元素 target
通过 flip(i + 1)
将翻到最前面,再进行 flip(target)
一次翻到正确位置即可。这样对每一个元素我们最多需要 2 次将其放在正确位置,因此总的次数最多为 2 * A.length
符合题目要求。此外针对每一个元素,首先寻找其在数组中的位置 O(n)
,第一次 flip O(n)
,第二次 flip O(n)
,所以总的时间复杂度为: O(n^2)
,代码如下所示:
#include <vector>
#include <algorithm>
class Solution {
public:
std::vector<int> pancakeSort(std::vector<int>& A) {
int n = A.size();
std::vector<int> results;
for (int target = n; target > 0; target--) {
int index;
// find current target from list
for (index = 0; index < target; index++) {
if (A[index] == target) break;
}
if (index == target - 1) continue;
// reverse [0, index] to get target to the first of list
if (index != 0) {
flip(A, index + 1);
results.push_back(index + 1);
}
// reverse [0, target - 1] to put target to correct position
if (index != target - 1){
flip(A, target);
results.push_back(target);
}
}
return results;
}
private:
// reverse the first k elements in the list
void flip(std::vector<int>& A, int k) {
std::reverse(A.begin(), A.begin() + k);
}
};
- 时间复杂度:O(n^2)
- 空间复杂度:O(1)