969. Pancake Sorting

Posted 2020-12-06 habibah-chang

问题：

给定数组，

假定反转动作k，表示：将数组前k个元素进行反转。

求反转动作k的序列，使得数组最终成为一个递增数组。（特：该数组为1～A.size()的一个排序）

Example 1:
Input: [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. 

Example 2:
Input: [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
A[i] is a permutation of [1, 2, ..., A.length]

　　

解法：

反转方法：

找到下一个最大的数nextmax的index：i

首先将这个数反转到开头，即A[0]的位置：即需要反转的k=i

然后将这个数反转到下一个最大位置nextmax-1：即需要反转的k=nextmax-1

这里要用到c++的反转数组函数reverse(A.begin(),A.end())->reverse(反转开头位置，反转结束位置+1)

因此，我们做一个元素的定位需要反转以下2次：

reverse(A.begin(),i+1);
reverse(A.begin(),nextmax);

 

代码参考：

 1 class Solution {
 2 public:
 3     vector<int> pancakeSort(vector<int>& A) {
 4         int i;
 5         int nextmax;
 6         vector<int> res;
 7         for(nextmax=A.size(); nextmax>0; nextmax--){
 8             for(i=0; A[i]!=nextmax; i++);
 9             if(i+1==nextmax) continue;//省略正反反转数相同的
10             reverse(A.begin(), A.begin()+i+1);
11             res.push_back(i+1);
12             reverse(A.begin(), A.begin()+nextmax);
13             res.push_back(nextmax);
14         }
15         return res;
16     }
17 };