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update: 添加问题“3264.K次乘运算后的最终数组I”的代码和题解
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Codes/3264-final-array-state-after-k-multiplication-operations-i.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,26 @@ | ||
| /* | ||
| * @Author: LetMeFly | ||
| * @Date: 2024-12-13 09:32:29 | ||
| * @LastEditors: LetMeFly.xyz | ||
| * @LastEditTime: 2024-12-13 09:33:35 | ||
| */ | ||
| #ifdef _WIN32 | ||
| #include "_[1,2]toVector.h" | ||
| #endif | ||
|
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| class Solution { | ||
| public: | ||
| vector<int> getFinalState(vector<int>& nums, int k, int multiplier) { | ||
| while (k--) { | ||
| int m = nums[0], loc = 0; | ||
| for (int i = 1; i < nums.size(); i++) { | ||
| if (nums[i] < m) { | ||
| m = nums[i]; | ||
| loc = i; | ||
| } | ||
| } | ||
| nums[loc] *= multiplier; | ||
| } | ||
| return nums; | ||
| } | ||
| }; |
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Codes/3264-final-array-state-after-k-multiplication-operations-i.go
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,20 @@ | ||
| /* | ||
| * @Author: LetMeFly | ||
| * @Date: 2024-12-13 09:40:31 | ||
| * @LastEditors: LetMeFly.xyz | ||
| * @LastEditTime: 2024-12-13 09:42:10 | ||
| */ | ||
| package main | ||
|
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| func getFinalState(nums []int, k int, multiplier int) []int { | ||
| for round := 0; round < k; round++ { | ||
| m, loc := nums[0], 0 | ||
| for i, val := range nums { | ||
| if val < m { | ||
| m, loc = val, i | ||
| } | ||
| } | ||
| nums[loc] *= multiplier | ||
| } | ||
| return nums | ||
| } |
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Codes/3264-final-array-state-after-k-multiplication-operations-i.java
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,21 @@ | ||
| /* | ||
| * @Author: LetMeFly | ||
| * @Date: 2024-12-13 09:37:28 | ||
| * @LastEditors: LetMeFly.xyz | ||
| * @LastEditTime: 2024-12-13 09:40:00 | ||
| */ | ||
| class Solution { | ||
| public int[] getFinalState(int[] nums, int k, int multiplier) { | ||
| for (int round = 0; round < k; round++) { | ||
| int m = nums[0], loc = 0; | ||
| for (int i = 1; i < nums.length; i++) { | ||
| if (nums[i] < m) { | ||
| m = nums[i]; | ||
| loc = i; | ||
| } | ||
| } | ||
| nums[loc] *= multiplier; | ||
| } | ||
| return nums; | ||
| } | ||
| } |
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Codes/3264-final-array-state-after-k-multiplication-operations-i.py
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| Original file line number | Diff line number | Diff line change |
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| ''' | ||
| Author: LetMeFly | ||
| Date: 2024-12-13 09:34:55 | ||
| LastEditors: LetMeFly.xyz | ||
| LastEditTime: 2024-12-13 09:36:26 | ||
| ''' | ||
| from typing import List | ||
|
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| class Solution: | ||
| def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]: | ||
| for _ in range(k): | ||
| m, loc = nums[0], 0 | ||
| for i, val in enumerate(nums): | ||
| if val < m: | ||
| m, loc = val, i | ||
| nums[loc] *= multiplier | ||
| return nums |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,198 @@ | ||
| --- | ||
| title: 3264.K 次乘运算后的最终数组 I | ||
| date: 2024-12-13 09:43:07 | ||
| tags: [题解, LeetCode, 简单, 数组, 数学, 模拟, 堆（优先队列）] | ||
| --- | ||
|
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| # 【LetMeFly】3264.K 次乘运算后的最终数组 I：模拟 | ||
|
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| 力扣题目链接：[https://leetcode.cn/problems/final-array-state-after-k-multiplication-operations-i/](https://leetcode.cn/problems/final-array-state-after-k-multiplication-operations-i/) | ||
|
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| <p>给你一个整数数组 <code>nums</code> ，一个整数 <code>k</code> 和一个整数 <code>multiplier</code> 。</p> | ||
|
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| <p>你需要对 <code>nums</code> 执行 <code>k</code> 次操作，每次操作中：</p> | ||
|
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| <ul> | ||
| <li>找到 <code>nums</code> 中的 <strong>最小</strong> 值 <code>x</code> ，如果存在多个最小值，选择最 <strong>前面</strong> 的一个。</li> | ||
| <li>将 <code>x</code> 替换为 <code>x * multiplier</code> 。</li> | ||
| </ul> | ||
|
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| <p>请你返回执行完 <code>k</code> 次乘运算之后，最终的 <code>nums</code> 数组。</p> | ||
|
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| <p> </p> | ||
|
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| <p><strong class="example">示例 1：</strong></p> | ||
|
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| <div class="example-block"> | ||
| <p><span class="example-io"><b>输入：</b>nums = [2,1,3,5,6], k = 5, multiplier = 2</span></p> | ||
|
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| <p><span class="example-io"><b>输出：</b>[8,4,6,5,6]</span></p> | ||
|
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| <p><strong>解释：</strong></p> | ||
|
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| <table> | ||
| <tbody> | ||
| <tr> | ||
| <th>操作</th> | ||
| <th>结果</th> | ||
| </tr> | ||
| <tr> | ||
| <td>1 次操作后</td> | ||
| <td>[2, 2, 3, 5, 6]</td> | ||
| </tr> | ||
| <tr> | ||
| <td>2 次操作后</td> | ||
| <td>[4, 2, 3, 5, 6]</td> | ||
| </tr> | ||
| <tr> | ||
| <td>3 次操作后</td> | ||
| <td>[4, 4, 3, 5, 6]</td> | ||
| </tr> | ||
| <tr> | ||
| <td>4 次操作后</td> | ||
| <td>[4, 4, 6, 5, 6]</td> | ||
| </tr> | ||
| <tr> | ||
| <td>5 次操作后</td> | ||
| <td>[8, 4, 6, 5, 6]</td> | ||
| </tr> | ||
| </tbody> | ||
| </table> | ||
| </div> | ||
|
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| <p><strong class="example">示例 2：</strong></p> | ||
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| <div class="example-block"> | ||
| <p><span class="example-io"><b>输入：</b></span>nums = [1,2], k = 3, multiplier = 4</p> | ||
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| <p><span class="example-io"><b>输出：</b></span>[16,8]</p> | ||
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| <p><strong>解释：</strong></p> | ||
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| <table> | ||
| <tbody> | ||
| <tr> | ||
| <th>操作</th> | ||
| <th>结果</th> | ||
| </tr> | ||
| <tr> | ||
| <td>1 次操作后</td> | ||
| <td>[4, 2]</td> | ||
| </tr> | ||
| <tr> | ||
| <td>2 次操作后</td> | ||
| <td>[4, 8]</td> | ||
| </tr> | ||
| <tr> | ||
| <td>3 次操作后</td> | ||
| <td>[16, 8]</td> | ||
| </tr> | ||
| </tbody> | ||
| </table> | ||
| </div> | ||
|
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| <p> </p> | ||
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| <p><strong>提示：</strong></p> | ||
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| <ul> | ||
| <li><code>1 <= nums.length <= 100</code></li> | ||
| <li><code>1 <= nums[i] <= 100</code></li> | ||
| <li><code>1 <= k <= 10</code></li> | ||
| <li><code>1 <= multiplier <= 5</code></li> | ||
| </ul> | ||
|
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| ## 解题方法：模拟 | ||
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| 进行$k$次如下操作： | ||
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| > 每次找到最小值所在位置，将最小值乘以$multiplier$ | ||
| 最后返回$nums$原数组。 | ||
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| + 时间复杂度$O(len(nums)\times k)$ | ||
| + 空间复杂度$O(1)$ | ||
|
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| ### AC代码 | ||
|
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| #### C++ | ||
|
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| ```cpp | ||
| class Solution { | ||
| public: | ||
| vector<int> getFinalState(vector<int>& nums, int k, int multiplier) { | ||
| while (k--) { | ||
| int m = nums[0], loc = 0; | ||
| for (int i = 1; i < nums.size(); i++) { | ||
| if (nums[i] < m) { | ||
| m = nums[i]; | ||
| loc = i; | ||
| } | ||
| } | ||
| nums[loc] *= multiplier; | ||
| } | ||
| return nums; | ||
| } | ||
| }; | ||
| ``` | ||
| #### Python | ||
| ```python | ||
| from typing import List | ||
| class Solution: | ||
| def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]: | ||
| for _ in range(k): | ||
| m, loc = nums[0], 0 | ||
| for i, val in enumerate(nums): | ||
| if val < m: | ||
| m, loc = val, i | ||
| nums[loc] *= multiplier | ||
| return nums | ||
| ``` | ||
|
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| #### Java | ||
|
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| ```java | ||
| class Solution { | ||
| public int[] getFinalState(int[] nums, int k, int multiplier) { | ||
| for (int round = 0; round < k; round++) { | ||
| int m = nums[0], loc = 0; | ||
| for (int i = 1; i < nums.length; i++) { | ||
| if (nums[i] < m) { | ||
| m = nums[i]; | ||
| loc = i; | ||
| } | ||
| } | ||
| nums[loc] *= multiplier; | ||
| } | ||
| return nums; | ||
| } | ||
| } | ||
| ``` | ||
|
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| #### Go | ||
|
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| ```go | ||
| package main | ||
|
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| func getFinalState(nums []int, k int, multiplier int) []int { | ||
| for round := 0; round < k; round++ { | ||
| m, loc := nums[0], 0 | ||
| for i, val := range nums { | ||
| if val < m { | ||
| m, loc = val, i | ||
| } | ||
| } | ||
| nums[loc] *= multiplier | ||
| } | ||
| return nums | ||
| } | ||
| ``` | ||
|
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| > 同步发文于CSDN和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2024/12/13/LeetCode%203264.K%E6%AC%A1%E4%B9%98%E8%BF%90%E7%AE%97%E5%90%8E%E7%9A%84%E6%9C%80%E7%BB%88%E6%95%B0%E7%BB%84I/)哦~ | ||
| > | ||
| > Tisfy：[https://letmefly.blog.csdn.net/article/details/144442552](https://letmefly.blog.csdn.net/article/details/144442552) |
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@@ -35,9 +35,17 @@ git clone --branch paper --single-branch [email protected]:LetMeFly666/SecFFT.git | |
| git reset --hard/--soft/--mixed HEAD^ | ||
| ``` | ||
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| <font color="red">下面先别看了，写的不对</font> | ||
| 其中： | ||
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| 其中`--hard`会强制变成上个版本（工作区暂存区清空），`--soft`会将相对上一个版本的变化放到暂存区，`--mixed`(默认选项)会将相对上一个版本的变化放到工作区。 | ||
| + `--hard`会强制变成上个版本（工作区暂存区清空） | ||
| + `--soft`会将相对上一个版本的变化保留到暂存区和工作区（已经add到暂存区过的变化还在暂存区 未add的变化还在工作区） | ||
| + `--mixed`(默认选项)会将相对上一个版本的变化全部放到工作区。 | ||
|
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| 另一个解释版本： | ||
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| + `--hard`：完全回退提交，丢弃暂存区和文件的所有修改。 | ||
| + `--soft`：回退提交，但保留文件和暂存区的修改。 | ||
| + `--mixed`：回退提交，丢弃暂存区的修改，但保留文件的修改。 | ||
|
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| 举个例子： | ||
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@@ -134,7 +142,7 @@ git reset --hard/--soft/--mixed HEAD^ | |
| > | ||
| > 注意Windows系统中`cmd`中的`^`大概是连接符的意思，可以使用`git reset --hard "HEAD^"`或`git reset --hard HEAD"^"`或`git reset --hard HEAD^^`来表示`HEAD^`。 | ||
| > | ||
| > 请注意，如果有未跟踪的内容（例如`echo 4 > 4`但是不`git add`），那么无论`git reset`时传递哪个参数，文件`4`都会原封不动地躺在工作区（这是因为历史记录中也没有文件`4`） TODO: 历史记录中存在文件4 | ||
| > 如果有**从**未跟踪的内容（例如`echo 4 > 4`但是不`git add`），那么无论`git reset`时传递哪个参数，文件`4`都会原封不动地躺在工作区（这是因为历史记录中也没有文件`4`） | ||
| #### 查看日志/commit记录 | ||
|
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