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198 changes: 198 additions & 0 deletions
solution/3600-3699/3687.Library Late Fee Calculator/README.md
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| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,198 @@ | ||
| --- | ||
| comments: true | ||
| difficulty: 简单 | ||
| edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3687.Library%20Late%20Fee%20Calculator/README.md | ||
| --- | ||
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| <!-- problem:start --> | ||
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| # [3687. Library Late Fee Calculator 🔒](https://leetcode.cn/problems/library-late-fee-calculator) | ||
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| [English Version](/solution/3600-3699/3687.Library%20Late%20Fee%20Calculator/README_EN.md) | ||
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| ## 题目描述 | ||
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| <!-- description:start --> | ||
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| <p>You are given an integer array <code>daysLate</code> where <code>daysLate[i]</code> indicates how many days late the <code>i<sup>th</sup></code> book was returned.</p> | ||
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| <p>The penalty is calculated as follows:</p> | ||
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| <ul> | ||
| <li>If <code>daysLate[i] == 1</code>, penalty is 1.</li> | ||
| <li>If <code>2 <= daysLate[i] <= 5</code>, penalty is <code>2 * daysLate[i]</code>.</li> | ||
| <li>If <code>daysLate[i] > 5</code>, penalty is <code>3 * daysLate[i]</code>.</li> | ||
| </ul> | ||
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| <p>Return the total penalty for all books.</p> | ||
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| <p> </p> | ||
| <p><strong class="example">Example 1:</strong></p> | ||
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| <div class="example-block"> | ||
| <p><strong>Input:</strong> <span class="example-io">daysLate = [5,1,7]</span></p> | ||
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| <p><strong>Output:</strong> <span class="example-io">32</span></p> | ||
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| <p><strong>Explanation:</strong></p> | ||
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| <ul> | ||
| <li><code>daysLate[0] = 5</code>: Penalty is <code>2 * daysLate[0] = 2 * 5 = 10</code>.</li> | ||
| <li><code>daysLate[1] = 1</code>: Penalty is <code>1</code>.</li> | ||
| <li><code>daysLate[2] = 7</code>: Penalty is <code>3 * daysLate[2] = 3 * 7 = 21</code>.</li> | ||
| <li>Thus, the total penalty is <code>10 + 1 + 21 = 32</code>.</li> | ||
| </ul> | ||
| </div> | ||
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| <p><strong class="example">Example 2:</strong></p> | ||
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| <div class="example-block"> | ||
| <p><strong>Input:</strong> <span class="example-io">daysLate = [1,1]</span></p> | ||
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| <p><strong>Output:</strong> <span class="example-io">2</span></p> | ||
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| <p><strong>Explanation:</strong></p> | ||
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| <ul> | ||
| <li><code>daysLate[0] = 1</code>: Penalty is <code>1</code>.</li> | ||
| <li><code>daysLate[1] = 1</code>: Penalty is <code>1</code>.</li> | ||
| <li>Thus, the total penalty is <code>1 + 1 = 2</code>.</li> | ||
| </ul> | ||
| </div> | ||
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| <p> </p> | ||
| <p><strong>Constraints:</strong></p> | ||
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| <ul> | ||
| <li><code>1 <= daysLate.length <= 100</code></li> | ||
| <li><code>1 <= daysLate[i] <= 100</code></li> | ||
| </ul> | ||
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| <!-- description:end --> | ||
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| ## 解法 | ||
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| <!-- solution:start --> | ||
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| ### 方法一:模拟 | ||
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| 我们定义一个函数 $\text{f}(x)$ 来计算每本书的罚款: | ||
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| $$ | ||
| \text{f}(x) = \begin{cases} | ||
| 1 & x = 1 \\ | ||
| 2x & 2 \leq x \leq 5 \\ | ||
| 3x & x > 5 | ||
| \end{cases} | ||
| $$ | ||
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| 然后我们对数组 $\textit{daysLate}$ 中的每个元素 $x$ 计算 $\text{f}(x)$ 并累加得到总罚款。 | ||
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| 时间复杂度 $O(n),ドル其中 $n$ 是数组 $\textit{daysLate}$ 的长度。空间复杂度 $O(1)$。 | ||
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| <!-- tabs:start --> | ||
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| #### Python3 | ||
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| ```python | ||
| class Solution: | ||
| def lateFee(self, daysLate: List[int]) -> int: | ||
| def f(x: int) -> int: | ||
| if x == 1: | ||
| return 1 | ||
| if x > 5: | ||
| return 3 * x | ||
| return 2 * x | ||
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| return sum(f(x) for x in daysLate) | ||
| ``` | ||
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| #### Java | ||
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| ```java | ||
| class Solution { | ||
| public int lateFee(int[] daysLate) { | ||
| IntUnaryOperator f = x -> { | ||
| if (x == 1) { | ||
| return 1; | ||
| } else if (x > 5) { | ||
| return 3 * x; | ||
| } else { | ||
| return 2 * x; | ||
| } | ||
| }; | ||
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| int ans = 0; | ||
| for (int x : daysLate) { | ||
| ans += f.applyAsInt(x); | ||
| } | ||
| return ans; | ||
| } | ||
| } | ||
| ``` | ||
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| #### C++ | ||
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| ```cpp | ||
| class Solution { | ||
| public: | ||
| int lateFee(vector<int>& daysLate) { | ||
| auto f = [](int x) { | ||
| if (x == 1) { | ||
| return 1; | ||
| } else if (x > 5) { | ||
| return 3 * x; | ||
| } else { | ||
| return 2 * x; | ||
| } | ||
| }; | ||
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| int ans = 0; | ||
| for (int x : daysLate) { | ||
| ans += f(x); | ||
| } | ||
| return ans; | ||
| } | ||
| }; | ||
| ``` | ||
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| #### Go | ||
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| ```go | ||
| func lateFee(daysLate []int) (ans int) { | ||
| f := func(x int) int { | ||
| if x == 1 { | ||
| return 1 | ||
| } else if x > 5 { | ||
| return 3 * x | ||
| } | ||
| return 2 * x | ||
| } | ||
| for _, x := range daysLate { | ||
| ans += f(x) | ||
| } | ||
| return | ||
| } | ||
| ``` | ||
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| #### TypeScript | ||
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| ```ts | ||
| function lateFee(daysLate: number[]): number { | ||
| const f = (x: number): number => { | ||
| if (x === 1) { | ||
| return 1; | ||
| } else if (x > 5) { | ||
| return 3 * x; | ||
| } | ||
| return 2 * x; | ||
| }; | ||
| return daysLate.reduce((acc, days) => acc + f(days), 0); | ||
| } | ||
| ``` | ||
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| <!-- tabs:end --> | ||
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| <!-- solution:end --> | ||
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| <!-- problem:end --> |
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