Question

969. Pancake Sorting

中文文档

Description

Given an array of integers arr, sort the array by performing a series of pancake flips.

In one pancake flip we do the following steps:

• Choose an integer k where 1 <= k <= arr.length.
• Reverse the sub-array arr[0...k-1] (0-indexed).

For example, if arr = [3,2,1,4] and we performed a pancake flip choosing k = 3, we reverse the sub-array [3,2,1], so arr = [1,2,3,4] after the pancake flip at k = 3.

Return _an array of the _k_-values corresponding to a sequence of pancake flips that sort _arr. Any valid answer that sorts the array within 10 * arr.length flips will be judged as correct.

 

Example 1:

Input: arr = [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: arr = [3, 2, 4, 1]
After 1st flip (k = 4): arr = [1, 4, 2, 3]
After 2nd flip (k = 2): arr = [4, 1, 2, 3]
After 3rd flip (k = 4): arr = [3, 2, 1, 4]
After 4th flip (k = 3): arr = [1, 2, 3, 4], which is sorted.

Example 2:

Input: arr = [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 <= arr.length <= 100
• 1 <= arr[i] <= arr.length
• All integers in arr are unique (i.e. arr is a permutation of the integers from 1 to arr.length).

Solutions

Solution 1

class Solution:
  def pancakeSort(self, arr: List[int]) -> List[int]:
    def reverse(arr, j):
      i = 0
      while i < j:
        arr[i], arr[j] = arr[j], arr[i]
        i, j = i + 1, j - 1

    n = len(arr)
    ans = []
    for i in range(n - 1, 0, -1):
      j = i
      while j > 0 and arr[j] != i + 1:
        j -= 1
      if j < i:
        if j > 0:
          ans.append(j + 1)
          reverse(arr, j)
        ans.append(i + 1)
        reverse(arr, i)
    return ans

class Solution {
  public List<Integer> pancakeSort(int[] arr) {
    int n = arr.length;
    List<Integer> ans = new ArrayList<>();
    for (int i = n - 1; i > 0; --i) {
      int j = i;
      for (; j > 0 && arr[j] != i + 1; --j)
        ;
      if (j < i) {
        if (j > 0) {
          ans.add(j + 1);
          reverse(arr, j);
        }
        ans.add(i + 1);
        reverse(arr, i);
      }
    }
    return ans;
  }

  private void reverse(int[] arr, int j) {
    for (int i = 0; i < j; ++i, --j) {
      int t = arr[i];
      arr[i] = arr[j];
      arr[j] = t;
    }
  }
}

class Solution {
public:
  vector<int> pancakeSort(vector<int>& arr) {
    int n = arr.size();
    vector<int> ans;
    for (int i = n - 1; i > 0; --i) {
      int j = i;
      for (; j > 0 && arr[j] != i + 1; --j)
        ;
      if (j == i) continue;
      if (j > 0) {
        ans.push_back(j + 1);
        reverse(arr.begin(), arr.begin() + j + 1);
      }
      ans.push_back(i + 1);
      reverse(arr.begin(), arr.begin() + i + 1);
    }
    return ans;
  }
};

func pancakeSort(arr []int) []int {
  var ans []int
  n := len(arr)
  reverse := func(j int) {
    for i := 0; i < j; i, j = i+1, j-1 {
      arr[i], arr[j] = arr[j], arr[i]
    }
  }
  for i := n - 1; i > 0; i-- {
    j := i
    for ; j > 0 && arr[j] != i+1; j-- {
    }
    if j < i {
      if j > 0 {
        ans = append(ans, j+1)
        reverse(j)
      }
      ans = append(ans, i+1)
      reverse(i)
    }
  }
  return ans
}

function pancakeSort(arr: number[]): number[] {
  let ans = [];
  for (let n = arr.length; n > 1; n--) {
    let index = 0;
    for (let i = 1; i < n; i++) {
      if (arr[i] >= arr[index]) {
        index = i;
      }
    }
    if (index == n - 1) continue;
    reverse(arr, index);
    reverse(arr, n - 1);
    ans.push(index + 1);
    ans.push(n);
  }
  return ans;
}

function reverse(nums: Array<number>, end: number): void {
  for (let i = 0, j = end; i < j; i++, j--) {
    [nums[i], nums[j]] = [nums[j], nums[i]];
  }
}

impl Solution {
  pub fn pancake_sort(mut arr: Vec<i32>) -> Vec<i32> {
    let mut res = vec![];
    for n in (1..arr.len()).rev() {
      let mut max_idx = 0;
      for idx in 0..=n {
        if arr[max_idx] < arr[idx] {
          max_idx = idx;
        }
      }
      if max_idx != n {
        if max_idx != 0 {
          arr[..=max_idx].reverse();
          res.push((max_idx as i32) + 1);
        }
        arr[..=n].reverse();
        res.push((n as i32) + 1);
      }
    }
    res
  }
}