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[Leetcode]969. Pancake Sorting (C++)

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 from 1 to A.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:

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#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)