update: 添加问题“3340.检查平衡字符串”的代码和题解(#809)

LetMeFly666 · LetMeFly666 · commit a79e876be6d6 · 2025-03-14T09:53:56.000+08:00
feat: #809 暂未验证，因为当前脚本使用的是 92df623 版 Signed-off-by: LetMeFly666 <814114971@qq.com>
diff --git a/.commitmsg b/.commitmsg
@@ -0,0 +1,3 @@
+
+feat: #809
+暂未验证，因为当前脚本使用的是 92df62367f78ad67cb781d7b4df8d7b27a272cbe 版
diff --git a/.nodeploy b/.nodeploy
diff --git a/Codes/3340-check-balanced-string.cpp b/Codes/3340-check-balanced-string.cpp
@@ -0,0 +1,20 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-03-14 09:30:43
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-03-14 09:32:51
+ */
+#ifdef _WIN32
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    bool isBalanced(string num) {
+        int cnt = 0;
+        for (int i = 0; i < num.size(); i++) {
+            cnt += i % 2 ? (num[i] - '0') : -(num[i] - '0');
+        }
+        return cnt == 0;
+    }
+};
diff --git a/Codes/3340-check-balanced-string.go b/Codes/3340-check-balanced-string.go
@@ -0,0 +1,19 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-03-14 09:36:55
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-03-14 09:38:01
+ */
+package main
+
+func isBalanced(num string) bool {
+    cnt := 0
+    for i, c := range num {
+        if i % 2 == 0 {
+            cnt += int(c) - 48
+        } else {
+            cnt -= int(c) - 48
+        }
+    }
+    return cnt == 0
+}
diff --git a/Codes/3340-check-balanced-string.java b/Codes/3340-check-balanced-string.java
@@ -0,0 +1,19 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-03-14 09:35:26
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-03-14 09:35:26
+ */
+class Solution {
+    public boolean isBalanced(String num) {
+        int cnt = 0;
+        for (int i = 0; i < num.length(); i++) {
+            if (i % 2 == 0) {
+                cnt += num.charAt(i) - 48;
+            } else {
+                cnt -= num.charAt(i) - 48;
+            }
+        }
+        return cnt == 0;
+    }
+}
diff --git a/Codes/3340-check-balanced-string.py b/Codes/3340-check-balanced-string.py
@@ -0,0 +1,12 @@
+'''
+Author: LetMeFly
+Date: 2025-03-14 09:34:04
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-03-14 09:34:04
+'''
+class Solution:
+    def isBalanced(self, num: str) -> bool:
+        cnt = 0
+        for i, c in enumerate(num):
+            cnt += ord(c) - 48 if i % 2 else 48 - ord(c)
+        return cnt == 0
diff --git a/README.md b/README.md
@@ -921,6 +921,7 @@
 |3297.统计重新排列后包含另一个字符串的子字符串数目I|中等|<a href="https://leetcode.cn/problems/count-substrings-that-can-be-rearranged-to-contain-a-string-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/01/09/LeetCode%203297.%E7%BB%9F%E8%AE%A1%E9%87%8D%E6%96%B0%E6%8E%92%E5%88%97%E5%90%8E%E5%8C%85%E5%90%AB%E5%8F%A6%E4%B8%80%E4%B8%AA%E5%AD%97%E7%AC%A6%E4%B8%B2%E7%9A%84%E5%AD%90%E5%AD%97%E7%AC%A6%E4%B8%B2%E6%95%B0%E7%9B%AEI/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/145031494" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-substrings-that-can-be-rearranged-to-contain-a-string-i/solutions/3042787/letmefly-3297tong-ji-zhong-xin-pai-lie-h-8ele/" target="_blank">LeetCode题解</a>|
 |3305.元音辅音字符串计数I|中等|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/03/12/LeetCode%203305.%E5%85%83%E9%9F%B3%E8%BE%85%E9%9F%B3%E5%AD%97%E7%AC%A6%E4%B8%B2%E8%AE%A1%E6%95%B0I/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146197747" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-i/solutions/3607571/letmefly-3305yuan-yin-fu-yin-zi-fu-chuan-ptkg/" target="_blank">LeetCode题解</a>|
 |3306.元音辅音字符串计数II|中等|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-ii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/03/13/LeetCode%203306.%E5%85%83%E9%9F%B3%E8%BE%85%E9%9F%B3%E5%AD%97%E7%AC%A6%E4%B8%B2%E8%AE%A1%E6%95%B0II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146228751" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-ii/solutions/3609545/letmefly-3306yuan-yin-fu-yin-zi-fu-chuan-7y2q/" target="_blank">LeetCode题解</a>|
+|3340.检查平衡字符串|简单|<a href="https://leetcode.cn/problems/check-balanced-string/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/03/14/LeetCode%203340.%E6%A3%80%E6%9F%A5%E5%B9%B3%E8%A1%A1%E5%AD%97%E7%AC%A6%E4%B8%B2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146249653" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/check-balanced-string/solutions/3610899/letmefly-3340jian-cha-ping-heng-zi-fu-ch-p8eo/" target="_blank">LeetCode题解</a>|
 |剑指Offer0047.礼物的最大价值|简单|<a href="https://leetcode.cn/problems/li-wu-de-zui-da-jie-zhi-lcof/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/03/08/LeetCode%20%E5%89%91%E6%8C%87%20Offer%2047.%20%E7%A4%BC%E7%89%A9%E7%9A%84%E6%9C%80%E5%A4%A7%E4%BB%B7%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/129408765" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/li-wu-de-zui-da-jie-zhi-lcof/solutions/2155672/letmefly-jian-zhi-offer-47li-wu-de-zui-d-rekb/" target="_blank">LeetCode题解</a>|
 |剑指OfferII0041.滑动窗口的平均值|简单|<a href="https://leetcode.cn/problems/qIsx9U/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/07/16/LeetCode%20%E5%89%91%E6%8C%87%20Offer%20II%200041.%20%E6%BB%91%E5%8A%A8%E7%AA%97%E5%8F%A3%E7%9A%84%E5%B9%B3%E5%9D%87%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/125819216" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/qIsx9U/solution/by-tisfy-30mq/" target="_blank">LeetCode题解</a>|
 |剑指OfferII0091.粉刷房子|中等|<a href="https://leetcode.cn/problems/JEj789/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/06/25/LeetCode%20%E5%89%91%E6%8C%87%20Offer%20II%200091.%20%E7%B2%89%E5%88%B7%E6%88%BF%E5%AD%90/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/125456885" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/JEj789/solution/letmefly-jian-zhi-offer-ii-091fen-shua-f-3olz/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 3340.检查平衡字符串.md b/Solutions/LeetCode 3340.检查平衡字符串.md
@@ -0,0 +1,156 @@
+---
+title: 3340.检查平衡字符串：模拟
+date: 2025-03-14 09:39:25
+tags: [题解, LeetCode, 简单, 字符串, 模拟]
+categories: [题解, LeetCode]
+---
+
+# 【LetMeFly】3340.检查平衡字符串：模拟
+
+力扣题目链接：[https://leetcode.cn/problems/check-balanced-string/](https://leetcode.cn/problems/check-balanced-string/)
+
+<p>给你一个仅由数字 0 - 9 组成的字符串 <code>num</code>。如果偶数下标处的数字之和等于奇数下标处的数字之和，则认为该数字字符串是一个 <b>平衡字符串</b>。</p>
+
+<p>如果 <code>num</code> 是一个 <strong>平衡字符串</strong>，则返回 <code>true</code>；否则，返回 <code>false</code>。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong>num<span class="example-io"> = "1234"</span></p>
+
+<p><strong>输出：</strong><span class="example-io">false</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>偶数下标处的数字之和为 <code>1 + 3 = 4</code>，奇数下标处的数字之和为 <code>2 + 4 = 6</code>。</li>
+	<li>由于 4 不等于 6，<code>num</code> 不是平衡字符串。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong>num<span class="example-io"> = "24123"</span></p>
+
+<p><strong>输出：</strong>true</p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>偶数下标处的数字之和为 <code>2 + 1 + 3 = 6</code>，奇数下标处的数字之和为 <code>4 + 2 = 6</code>。</li>
+	<li>由于两者相等，<code>num</code> 是平衡字符串。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= num.length &lt;= 100</code></li>
+	<li><code>num</code> 仅由数字 0 - 9 组成。</li>
+</ul>
+
+
+    
+## 解题方法：遍历求和
+
+使用一个整型变量$cnt$来统计结果即可。遍历字符串，遇到奇数下标则加上当前字符对应的数字，否则减去之。最终判断$cnt$是否为$0$。
+
++ 时间复杂度$O(len(num))$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-03-14 09:30:43
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-03-14 09:32:51
+ */
+class Solution {
+public:
+    bool isBalanced(string num) {
+        int cnt = 0;
+        for (int i = 0; i < num.size(); i++) {
+            cnt += i % 2 ? (num[i] - '0') : -(num[i] - '0');
+        }
+        return cnt == 0;
+    }
+};
+```
+
+#### Python
+
+```python
+'''
+Author: LetMeFly
+Date: 2025-03-14 09:34:04
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-03-14 09:34:04
+'''
+class Solution:
+    def isBalanced(self, num: str) -> bool:
+        cnt = 0
+        for i, c in enumerate(num):
+            cnt += ord(c) - 48 if i % 2 else 48 - ord(c)
+        return cnt == 0
+```
+
+#### Java
+
+```java
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-03-14 09:35:26
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-03-14 09:35:26
+ */
+class Solution {
+    public boolean isBalanced(String num) {
+        int cnt = 0;
+        for (int i = 0; i < num.length(); i++) {
+            if (i % 2 == 0) {
+                cnt += num.charAt(i) - 48;
+            } else {
+                cnt -= num.charAt(i) - 48;
+            }
+        }
+        return cnt == 0;
+    }
+}
+```
+
+#### Go
+
+```go
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-03-14 09:36:55
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-03-14 09:38:01
+ */
+package main
+
+func isBalanced(num string) bool {
+    cnt := 0
+    for i, c := range num {
+        if i % 2 == 0 {
+            cnt += int(c) - 48
+        } else {
+            cnt -= int(c) - 48
+        }
+    }
+    return cnt == 0
+}
+```
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/146249653)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/2025/03/14/LeetCode%203340.%E6%A3%80%E6%9F%A5%E5%B9%B3%E8%A1%A1%E5%AD%97%E7%AC%A6%E4%B8%B2/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
diff --git a/categoryWhenNewSolution.py b/categoryWhenNewSolution.py
@@ -0,0 +1,32 @@
+'''
+Author: LetMeFly
+Date: 2025-03-14 09:49:08
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-03-14 09:51:50
+Command: python categoryWhenNewSolution.py 3340. 检查平衡字符串
+'''
+import sys
+import time
+
+
+argv = sys.argv
+num = int(argv[1][:-1])
+title = ""
+for i in range(2, len(argv)):
+    if i != 2:
+        title += " "
+    title += argv[i]
+nameProblem = "AllProblems/{0}.{1}.md".format(num, title)
+
+with open(nameProblem, "r", encoding="utf-8") as f:
+    problem = f.read()
+
+solution = problem
+def refreshPublistTime(solution: str) -> str:
+    splited = solution.split("\n")
+    splited[2] = "date: " + time.strftime("%Y-%m-%d %H:%M:%S", time.localtime())
+    splited.insert(4, 'categories: [题解, LeetCode]')
+    return "\n".join(splited)
+
+solution = refreshPublistTime(solution)
+print(solution)
diff --git a/newSolution.py b/newSolution.py
@@ -2,7 +2,7 @@
 Author: LetMeFly
 Date: 2022-07-03 11:21:14
 LastEditors: LetMeFly.xyz
-LastEditTime: 2025-03-12 10:40:29
+LastEditTime: 2025-03-14 09:49:25
 Command: python newSolution.py 102. 二叉树的层序遍历
 What's more: 当前仅支持数字开头的题目
 What's more: 代码结构写的很混乱 - 想单文件实现所有操作
@@ -126,6 +126,7 @@ def genSolutionPart(num):
 def refreshPublistTime(solution: str) -> str:
     splited = solution.split("\n")
     splited[2] = "date: " + time.strftime("%Y-%m-%d %H:%M:%S", time.localtime())
+    splited.insert(4, 'categories: [题解, LeetCode]')
     return "\n".join(splited)
 
 solution = refreshPublistTime(solution)

Original file line number	Diff line number	Diff line change
`@@ -0,0 +1,3 @@`
	`1`	`+`
	`2`	`+feat: #809`
	`3`	`+暂未验证，因为当前脚本使用的是 92df62367f78ad67cb781d7b4df8d7b27a272cbe 版`