leetcode 2140. Solving Questions With Brainpower（python）
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前言
由亚马逊公司赞助的 Leetcode Weekly Contest 276 ，优秀者还能获得亚马逊公司的面试机会（慕了），看了一下榜单第一名是位北大的选手，为国争光，只要第一名国籍是中国的就好，吾辈楷模，我的水平令人惭愧。本文介绍的是 276 周赛第三道题目，难度 Medium ，耗时一个小时，感觉也不是很难，但是就感觉差那么一点点，没做出来，考察的是动态规划，但是我用递归解题超时了，最后也没优化出来。
描述
You are given a 0-indexed 2D integer array questions where questions[i] = [pointsi, brainpoweri].
The array describes the questions of an exam, where you have to process the questions in order (i.e., starting from question 0) and make a decision whether to solve or skip each question. Solving question i will earn you pointsi points but you will be unable to solve each of the next brainpoweri questions. If you skip question i, you get to make the decision on the next question.
- For example, given questions = [[3, 2], [4, 3], [4, 4], [2, 5]]:
- If question 0 is solved, you will earn 3 points but you will be unable to solve questions 1 and 2.
- If instead, question 0 is skipped and question 1 is solved, you will earn 4 points but you will be unable to solve questions 2 and 3.
Return the maximum points you can earn for the exam.
Example 1:
Input: questions = [[3,2],[4,3],[4,4],[2,5]]
Output: 5
Explanation: The maximum points can be earned by solving questions 0 and 3.
- Solve question 0: Earn 3 points, will be unable to solve the next 2 questions
- Unable to solve questions 1 and 2
- Solve question 3: Earn 2 points
Total points earned: 3 + 2 = 5. There is no other way to earn 5 or more points.
Example 2:
Input: questions = [[1,1],[2,2],[3,3],[4,4],[5,5]]
Output: 7
Explanation: The maximum points can be earned by solving questions 1 and 4.
- Skip question 0
- Solve question 1: Earn 2 points, will be unable to solve the next 2 questions
- Unable to solve questions 2 and 3
- Solve question 4: Earn 5 points
Total points earned: 2 + 5 = 7. There is no other way to earn 7 or more points.
Note:
1 <= questions.length <= 105
questions[i].length == 2
1 <= pointsi, brainpoweri <= 105
解析
根据题意，有一个 0 索引的 2D 整数数组 questions ，其中 questions[i] = [pointsi, brainpoweri]。该数组描述了考试的问题，必须按顺序从左到右处理问题，但是对于每个问题可以决定是解决还是跳过。 如果选择解决 questions[i] 将赢得 pointsi 个积分，但将无法解决接下来的 brainpoweri 问题。 如果你跳过第 i 题，你就可以去处理下一个问题。
- 例如，给定 questions = [[3, 2], [4, 3], [4, 4], [2, 5]] ：
- 如果解决了第 0 题，将获得 3 分，但将无法解决第 1 题和第 2 题
- 如果跳过第 0 题去解决第 1 题，您将获得 4 分，但您将无法解决第 2 题和第 3 题。
最后返回通过考试获得的最高分数。
- 首先我先用 pos 去找到如果解决第 i 个问题，下一个可以解决的问题的索引，
- 然后使用 DFS 函数，函数表示从 i 个位置开始解决问题，可以得到的最大的分值
- 遍历所有问题的位置，计算 dfs(j, questions[j][0]) ，得到最后的最大的分值
我的解法是使用了递归 ，但是超时了，因为使用了 DFS 和两次循环，导致时间复杂度成了 O(n^3) 。
解答
class Solution(object):
def mostPoints(self, questions):
"""
:type questions: List[List[int]]
:rtype: int
"""
pos = []
N = len(questions)
for i,(coin, brainpower) in enumerate(questions):
pos.append(min(i+brainpower+1, N))
def dfs(i, coins):
if i >= N - 1 or pos[i]>=N:
return coins
return max([dfs(j, coins + questions[j][0]) for j in range(pos[i], N)])
return max([dfs(j, questions[j][0]) for j in range(N)])
运行结果
Time Limit Exceeded
解析
看了其他大佬的解法，我发现我的思路应该变一下，不应该从前到后，而是从后向前，对于每道题目我们都知道有两种方案，一种是跳过去执行下一个题目，另一种是解决当前的问题而跳过后面若干到题目，要求我们最后得到的分数最大，这是典型的动态规划题目，我们使用动态规划的思路，从后往前遍历 questions ，定义 dp[i] 为 questions[i:] 这个子字符串能收集到的最大分数，长度为 N+1 ，因为我们要使用 dp[N] 来保存超过 questions 长度的索引的值为 0 ，动态规划的公式为
dp[i] = max(points + dp[min(jump + i + 1, len(questions))], dp[i + 1])
，从后往前遍历结束之后，返回 dp[0] 即可。返回 dp[0] 即可。时间复杂度是 O(n) ，空间复杂度是 O(n) 。
解答
class Solution(object):
def mostPoints(self, questions):
"""
:type questions: List[List[int]]
:rtype: int
"""
dp = [0] * (len(questions) + 1)
for i in range(len(questions) - 1, -1, -1):
points, jump = questions[i][0], questions[i][1]
dp[i] = max(points + dp[min(jump + i + 1, len(questions))], dp[i + 1])
return dp[0];
运行结果
Runtime: 2241 ms
Memory Usage: 65.5 MB
转载自:https://juejin.cn/post/7055085945023889415