Pancake Sorting (M)
题目
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
kwhere0 <= 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.
Example 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.
Example 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.
Constraints:
1 <= A.length <= 1001 <= A[i] <= A.length- All integers in
Aare unique (i.e.Ais a permutation of the integers from1toA.length).
题意
只使用翻转操作(每次可翻转[0...k])来排序一个数组。注意k实际上表示翻转的数量。
思路
先找到最大值n，执行两次翻转，先翻转到第一位，再翻转到最后一位，这样n就已经在它应在的位置上了。再找到n-1，将其翻转到倒数第二位。以此类推。
代码实现
Java
class Solution {
public List<Integer> pancakeSort(int[] A) {
List<Integer> ans = new ArrayList<>();
for (int i = A.length; i >= 2; i--) {
for (int j = 0; j < i - 1; j++) {
if (A[j] == i) {
ans.add(j + 1);
flip(A, 0, j);
ans.add(i);
flip(A, 0, i - 1);
break;
}
}
}
return ans;
}
private void flip(int[] A, int i, int j) {
while (i < j) {
int tmp = A[i];
A[i++] = A[j];
A[j--] = tmp;
}
}
}
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