添加了问题“2578. 最小和分割”的代码和题解

Author: LetMeFly666

Codes/2578-split-with-minimum-sum.cpp

@@ -0,0 +1,23 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2023-10-09 09:06:16
+ * @LastEditors: LetMeFly
+ * @LastEditTime: 2023-10-09 09:09:44
+ */
+#ifdef _WIN32
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    int splitNum(int num) {
+        string s = to_string(num);
+        sort(s.begin(), s.end());
+        string n1, n2;
+        for (int i = 0; i < s.size(); i++) {
+            (i % 2 ? n2 : n1) += s[i];
+        }
+        // cout << "n1: " << n1 << ", n2: " << n2 << endl;  //**********
+        return atoi(n1.c_str()) + atoi(n2.c_str());
+    }
+};

Codes/2578-split-with-minimum-sum.py

@@ -0,0 +1,13 @@
+'''
+Author: LetMeFly
+Date: 2023-10-09 09:10:33
+LastEditors: LetMeFly
+LastEditTime: 2023-10-09 09:12:51
+'''
+class Solution:
+    def splitNum(self, num: int) -> int:
+        s = sorted(str(num))
+        n = ['', '']
+        for i in range(len(s)):
+            n[i % 2] += s[i]
+        return int(n[0]) + int(n[1])

README.md

@@ -507,6 +507,7 @@ int main() {
 |2511.最多可以摧毁的敌人城堡数目|简单|<a href="https://leetcode.cn/problems/maximum-enemy-forts-that-can-be-captured/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2023/09/02/LeetCode%202511.%E6%9C%80%E5%A4%9A%E5%8F%AF%E4%BB%A5%E6%91%A7%E6%AF%81%E7%9A%84%E6%95%8C%E4%BA%BA%E5%9F%8E%E5%A0%A1%E6%95%B0%E7%9B%AE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/132634912" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-enemy-forts-that-can-be-captured/solutions/2422392/letmefly-2511zui-duo-ke-yi-cui-hui-de-di-r7kt/" target="_blank">LeetCode题解</a>|
 |2544.交替数字和|简单|<a href="https://leetcode.cn/problems/alternating-digit-sum/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2023/07/12/LeetCode%202544.%E4%BA%A4%E6%9B%BF%E6%95%B0%E5%AD%97%E5%92%8C/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131673485" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/alternating-digit-sum/solutions/2340450/letmefly-2544jiao-ti-shu-zi-he-by-tisfy-aa0f/" target="_blank">LeetCode题解</a>|
 |2559.统计范围内的元音字符串数|中等|<a href="https://leetcode.cn/problems/count-vowel-strings-in-ranges/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2023/06/02/LeetCode%202559.%E7%BB%9F%E8%AE%A1%E8%8C%83%E5%9B%B4%E5%86%85%E7%9A%84%E5%85%83%E9%9F%B3%E5%AD%97%E7%AC%A6%E4%B8%B2%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131014779" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-vowel-strings-in-ranges/solutions/2294361/letmefly-2559tong-ji-fan-wei-nei-de-yuan-mq4f/" target="_blank">LeetCode题解</a>|
+|2578.最小和分割|简单|<a href="https://leetcode.cn/problems/split-with-minimum-sum/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2023/10/09/LeetCode%202578.%E6%9C%80%E5%B0%8F%E5%92%8C%E5%88%86%E5%89%B2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/133694187" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/split-with-minimum-sum/solutions/2473429/letmefly-2578zui-xiao-he-fen-ge-tan-xin-pxbyl/" target="_blank">LeetCode题解</a>|
 |2582.递枕头|简单|<a href="https://leetcode.cn/problems/pass-the-pillow/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2023/09/26/LeetCode%202582.%E9%80%92%E6%9E%95%E5%A4%B4/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/133294825" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/pass-the-pillow/solutions/2459194/letmefly-2582di-zhen-tou-qing-xi-de-hua-ienu5/" target="_blank">LeetCode题解</a>|
 |2591.将钱分给最多的儿童|简单|<a href="https://leetcode.cn/problems/distribute-money-to-maximum-children/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2023/09/22/LeetCode%202591.%E5%B0%86%E9%92%B1%E5%88%86%E7%BB%99%E6%9C%80%E5%A4%9A%E7%9A%84%E5%84%BF%E7%AB%A5/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/133167460" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/distribute-money-to-maximum-children/solutions/2454112/letmefly-2591jiang-qian-fen-gei-zui-duo-2k1tq/" target="_blank">LeetCode题解</a>|
 |2596.检查骑士巡视方案|中等|<a href="https://leetcode.cn/problems/check-knight-tour-configuration/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2023/09/13/LeetCode%202596.%E6%A3%80%E6%9F%A5%E9%AA%91%E5%A3%AB%E5%B7%A1%E8%A7%86%E6%96%B9%E6%A1%88/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/132847346" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/check-knight-tour-configuration/solutions/2440047/letmefly-2596jian-cha-qi-shi-xun-shi-fan-2q5j/" target="_blank">LeetCode题解</a>|

Solutions/LeetCode 2578.最小和分割.md

@@ -0,0 +1,108 @@
+---
+title: 2578.最小和分割
+date: 2023-10-09 09:13:42
+tags: [题解, LeetCode, 简单, 贪心, 数学, 排序]
+---
+
+# 【LetMeFly】2578.最小和分割：贪心（数学）
+
+力扣题目链接：[https://leetcode.cn/problems/split-with-minimum-sum/](https://leetcode.cn/problems/split-with-minimum-sum/)
+
+<p>给你一个正整数&nbsp;<code>num</code>&nbsp;，请你将它分割成两个非负整数&nbsp;<code>num1</code> 和&nbsp;<code>num2</code>&nbsp;，满足：</p>
+
+<ul>
+	<li><code>num1</code> 和&nbsp;<code>num2</code>&nbsp;直接连起来，得到&nbsp;<code>num</code>&nbsp;各数位的一个排列。
+
+	<ul>
+		<li>换句话说，<code>num1</code> 和&nbsp;<code>num2</code>&nbsp;中所有数字出现的次数之和等于&nbsp;<code>num</code>&nbsp;中所有数字出现的次数。</li>
+	</ul>
+	</li>
+	<li><code>num1</code> 和&nbsp;<code>num2</code>&nbsp;可以包含前导 0 。</li>
+</ul>
+
+<p>请你返回&nbsp;<code>num1</code> 和 <code>num2</code>&nbsp;可以得到的和的 <strong>最小</strong> 值。</p>
+
+<p><strong>注意：</strong></p>
+
+<ul>
+	<li><code>num</code>&nbsp;保证没有前导 0 。</li>
+	<li><code>num1</code> 和&nbsp;<code>num2</code>&nbsp;中数位顺序可以与&nbsp;<code>num</code>&nbsp;中数位顺序不同。</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>num = 4325
+<b>输出：</b>59
+<b>解释：</b>我们可以将 4325 分割成 <code>num1 </code>= 24 和 num2<code> = </code>35 ，和为 59 ，59 是最小和。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>num = 687
+<b>输出：</b>75
+<b>解释：</b>我们可以将 687 分割成 <code>num1</code> = 68 和 <code>num2 </code>= 7 ，和为最优值 75 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>10 &lt;= num &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+
+    
+## 方法一：贪心（数学）
+
+先说结论：将给定数字转为字符串后将其中字符从小到大排序，然后依次分配给两个新数字即可。
+
+不严谨的原理描述：
+
++ 越高位数字尽量越小，因此要从小到大排序
++ 最终返回的是两数之和，所以首先位数越小越好，因此尽可能两个数字长度相等
++ 若两个数长度不相等，更长的那个数字的最高位要尽可能小（例如将```23456```分成```246```和```35```，唯一的百位是最小的```2```）
+
+结论中描述的分法恰好满足。
+
++ 时间复杂度$O(k\log k)$，其中$k = \log num$
++ 空间复杂度$O(\log k)$
+
+### AC代码
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int splitNum(int num) {
+        string s = to_string(num);
+        sort(s.begin(), s.end());
+        string n1, n2;
+        for (int i = 0; i < s.size(); i++) {
+            (i % 2 ? n2 : n1) += s[i];
+        }
+        // cout << "n1: " << n1 << ", n2: " << n2 << endl;  //**********
+        return atoi(n1.c_str()) + atoi(n2.c_str());
+    }
+};
+```
+
+#### Python
+
+```python
+class Solution:
+    def splitNum(self, num: int) -> int:
+        s = sorted(str(num))
+        n = ['', '']
+        for i in range(len(s)):
+            n[i % 2] += s[i]
+        return int(n[0]) + int(n[1])
+```
+
+> 同步发文于CSDN，原创不易，转载经作者同意后请附上[原文链接](https://blog.tisfy.eu.org/2023/10/09/LeetCode%202578.%E6%9C%80%E5%B0%8F%E5%92%8C%E5%88%86%E5%89%B2/)哦~
+> Tisfy：[https://letmefly.blog.csdn.net/article/details/133694187](https://letmefly.blog.csdn.net/article/details/133694187)