118th LeetCode Weekly Contest Pancake Sorting

Given an array A, we can perform a pancake flip: We choose some positive integer k <= A.length, then reverse the order of the first kelements of A.  We want to perform zero or more pancake flips (doing them one after another in succession) to sort the array A.

Return the k-values corresponding to a sequence of pancake flips that sort A.  Any valid answer that sorts the array within 10 * A.lengthflips will be judged as correct.

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. 1 <= A.length <= 100
2. A[i] is a permutation of [1, 2, ..., A.length]

对于这种题倒着处理。

主要是看两种情况，

3 1 2

要把2放后面，这1 2 3变成2 1 3，再全部倒着来一次3 1 2就OK

2 3 1这种，就直接全部倒着来就行。

class Solution {
public:
    int arr[400];
    void reverse1(int from,int to) {
            while(from<to){
                int tmp=arr[from];
                arr[from++]=arr[to];
                arr[to--]=tmp;
            }
    }
    int Find(int num,int last){
        for(int i=0;i<last;i++){
            if(arr[i]==num){
                return i+1;
            }
        }
        return 0;
    }
    void x(int len){
        for(int i=0;i<len;i++){
            cout<<arr[i]<<" ";
        }
        cout<<endl;
    }

    vector<int> pancakeSort(vector<int>& A) {
        vector<int>Ve = A;
        vector<int>Vee;
        sort(Ve.begin(),Ve.end());

        int Size = A.size();

        for(int i=0;i<Size;i++){
            arr[i] = Ve[i];
        }

        for(int i=0;i<Size;i++){
           int address = Find(A[Size - i - 1],Size-i);
           if(address == 1){
                //cout<<address<<" A"<<endl;
                reverse1(0,Size - i - 1);
                //x(Size);
                if(Size - i == 1){
                    continue;
                }
                Vee.push_back(Size - i);

           }else if(address == Size - i){
                //cout<<address<<" B "<<Size - i<<endl;
                continue;
           }else{
               //cout<<address<<" C"<<endl;
                reverse1(0,address - 1);
                //x(Size);
               if(address == 1){
                    continue;
                }
                Vee.push_back(address);
                reverse1(0,Size - i - 1);
                //x(Size);
               if(Size - i == 1){
                    continue;
                }
                Vee.push_back(Size - i);
           }
        }
        reverse(Vee.begin(),Vee.end());
        return Vee;
    }
};

原文地址：https://www.cnblogs.com/yinghualuowu/p/10230810.html