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.length
flips 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 <= A.length <= 100
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[];
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=;i<last;i++){
if(arr[i]==num){
return i+;
}
}
return ;
}
void x(int len){
for(int i=;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=;i<Size;i++){
arr[i] = Ve[i];
} for(int i=;i<Size;i++){
int address = Find(A[Size - i - ],Size-i);
if(address == ){
//cout<<address<<" A"<<endl;
reverse1(,Size - i - );
//x(Size);
if(Size - i == ){
continue;
}
Vee.push_back(Size - i); }else if(address == Size - i){
//cout<<address<<" B "<<Size - i<<endl;
continue;
}else{
//cout<<address<<" C"<<endl;
reverse1(,address - );
//x(Size);
if(address == ){
continue;
}
Vee.push_back(address);
reverse1(,Size - i - );
//x(Size);
if(Size - i == ){
continue;
}
Vee.push_back(Size - i);
}
}
reverse(Vee.begin(),Vee.end());
return Vee;
}
};
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