Question

16.11. Diving Board

中文文档

Description

You are building a diving board by placing a bunch of planks of wood end-to-end. There are two types of planks, one of length shorter and one of length longer. You must use exactly K planks of wood. Write a method to generate all possible lengths for the diving board.

return all lengths in non-decreasing order.

Example:

Input: 

shorter = 1

longer = 2

k = 3

Output:  {3,4,5,6}

Note:

• 0 < shorter <= longer
• 0 <= k <= 100000

Solutions

Solution 1: Case Analysis

If $k=0$, there is no solution, and we can directly return an empty list.

If $shorter=longer$, we can only use a board with length $longer \times k$, so we directly return a list with length $longer \times k$.

Otherwise, we can use a board with length $shorter \times (k-i) + longer \times i$, where $0 \leq i \leq k$. We enumerate $i$ in the range $[0, k]$, and calculate the corresponding length. For different values of $i$, we will not get the same length, because if $0 \leq i \lt j \leq k$, then the difference in length is $(i - j) \times (longer - shorter) \lt 0$. Therefore, for different values of $i$, we will get different lengths.

The time complexity is $O(k)$, where $k$ is the number of boards. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

class Solution:
  def divingBoard(self, shorter: int, longer: int, k: int) -> List[int]:
    if k == 0:
      return []
    if longer == shorter:
      return [longer * k]
    ans = []
    for i in range(k + 1):
      ans.append(longer * i + shorter * (k - i))
    return ans

class Solution {
  public int[] divingBoard(int shorter, int longer, int k) {
    if (k == 0) {
      return new int[0];
    }
    if (longer == shorter) {
      return new int[] {longer * k};
    }
    int[] ans = new int[k + 1];
    for (int i = 0; i < k + 1; ++i) {
      ans[i] = longer * i + shorter * (k - i);
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> divingBoard(int shorter, int longer, int k) {
    if (k == 0) return {};
    if (longer == shorter) return {longer * k};
    vector<int> ans;
    for (int i = 0; i < k + 1; ++i)
      ans.push_back(longer * i + shorter * (k - i));
    return ans;
  }
};

func divingBoard(shorter int, longer int, k int) []int {
  if k == 0 {
    return []int{}
  }
  if longer == shorter {
    return []int{longer * k}
  }
  var ans []int
  for i := 0; i < k+1; i++ {
    ans = append(ans, longer*i+shorter*(k-i))
  }
  return ans
}

function divingBoard(shorter: number, longer: number, k: number): number[] {
  if (k === 0) {
    return [];
  }
  if (longer === shorter) {
    return [longer * k];
  }
  const ans: number[] = [k + 1];
  for (let i = 0; i <= k; ++i) {
    ans[i] = longer * i + shorter * (k - i);
  }
  return ans;
}