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 k elements 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]
代码
Approach #1
Explanation Find the index i of the next maximum number x. Reverse i + 1 numbers, so that the x will be at A[0] Reverse x numbers, so that x will be at A[x - 1]. Repeat this process N times.
Update: Actually, I didn't use the condition permutation of [1,2,..., A.length]. I searched in the descending order of A.
Time Complexity: O(N^2)
class Solution {
public List<Integer> pancakeSort(int[] A) {
List<Integer> res = new ArrayList<>();
for (int x = A.length, i; x > 0; --x) {
for (i = 0; A[i] != x; ++i);
reverse(A, i + 1); // 把最大的值移到A[0]
res.add(i + 1);
reverse(A, x); // 把最大的移到A[end]
res.add(x);
}
return res;
}
public void reverse(int[] A, int k) {
for (int i = 0, j = k - 1; i < j; i++, j--) {
int tmp = A[i];
A[i] = A[j];
A[j] = tmp;
}
}
}
Approach #2 Sort Largest to Smallest
Complexity Analysis
Time Complexity: O(N^2), where NN is the length of A.
Space Complexity: O(N).
class Solution {
public List<Integer> pancakeSort(int[] A) {
List<Integer> ans = new ArrayList();
int N = A.length;
Integer[] B = new Integer[N];
for (int i = 0; i < N; i++) {
B[i] = i + 1；
}
Arrays.sort(B, (i, j) -> A[j - 1] - A[i - 1]);
for (int i: B) {
for (int f: ans) {
if (i <= f) {
i = f + 1 - i;
}
}
ans.add(i);
ans.add(N--);
}
return ans;
}
}