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feat: add solutions to lc problems: No.2678~2681
* No.2678.Number of Senior Citizens * No.2679.Sum in a Matrix * No.2680.Maximum OR * No.2681.Power of Heroes
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‎solution/2600-2699/2678.Number of Senior Citizens/README.md

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# [2678. 老人的数目](https://leetcode.cn/problems/number-of-senior-citizens)
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[English Version](/solution/2600-2699/2678.Number%20of%20Senior%20Citizens/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个下标从 <strong>0</strong>&nbsp;开始的字符串&nbsp;<code>details</code>&nbsp;。<code>details</code>&nbsp;中每个元素都是一位乘客的信息，信息用长度为 <code>15</code>&nbsp;的字符串表示，表示方式如下：</p>
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<ul>
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<li>前十个字符是乘客的手机号码。</li>
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<li>接下来的一个字符是乘客的性别。</li>
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<li>接下来两个字符是乘客的年龄。</li>
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<li>最后两个字符是乘客的座位号。</li>
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</ul>
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<p>请你返回乘客中年龄 <strong>严格大于 60 岁</strong>&nbsp;的人数。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<b>输入：</b>details = ["7868190130M7522","5303914400F9211","9273338290F4010"]
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<b>输出：</b>2
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<b>解释：</b>下标为 0 ，1 和 2 的乘客年龄分别为 75 ，92 和 40 。所以有 2 人年龄大于 60 岁。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<b>输入：</b>details = ["1313579440F2036","2921522980M5644"]
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<b>输出：</b>0
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<b>解释：</b>没有乘客的年龄大于 60 岁。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= details.length &lt;= 100</code></li>
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<li><code>details[i].length == 15</code></li>
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<li><code>details[i]</code>&nbsp;中的数字只包含&nbsp;<code>'0'</code>&nbsp;到&nbsp;<code>'9'</code>&nbsp;。</li>
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<li><code>details[i][10]</code>&nbsp;是 <code>'M'</code>&nbsp;，<code>'F'</code>&nbsp;或者&nbsp;<code>'O'</code>&nbsp;之一。</li>
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<li>所有乘客的手机号码和座位号互不相同。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：遍历计数**
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我们可以遍历 `details` 中的每个字符串 $x$，并将 $x$ 的第 $12$ 和第 $13$ 个字符（下标为 $11$, $12$）转换为整数，判断是否大于 $60$，如果是则将答案加一。
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遍历结束后，返回答案即可。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 是 `details` 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def countSeniors(self, details: List[str]) -> int:
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return sum(int(x[11:13]) > 60 for x in details)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int countSeniors(String[] details) {
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int ans = 0;
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for (var x : details) {
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int age = Integer.parseInt(x.substring(11, 13));
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if (age > 60) {
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++ans;
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countSeniors(vector<string>& details) {
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int ans = 0;
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for (auto& x : details) {
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int age = stoi(x.substr(11, 2));
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ans += age > 60;
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func countSeniors(details []string) (ans int) {
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for _, x := range details {
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age, _ := strconv.Atoi(x[11:13])
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if age > 60 {
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ans++
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}
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}
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return
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2600-2699/2678.Number of Senior Citizens/README_EN.md

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# [2678. Number of Senior Citizens](https://leetcode.com/problems/number-of-senior-citizens)
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[中文文档](/solution/2600-2699/2678.Number%20of%20Senior%20Citizens/README.md)
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## Description
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<p>You are given a <strong>0-indexed</strong> array of strings <code>details</code>. Each element of <code>details</code> provides information about a given passenger compressed into a string of length <code>15</code>. The system is such that:</p>
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<ul>
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<li>The first ten characters consist of the phone number of passengers.</li>
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<li>The next character denotes the gender of the person.</li>
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<li>The following two characters are used to indicate the age of the person.</li>
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<li>The last two characters determine the seat allotted to that person.</li>
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</ul>
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<p>Return <em>the number of passengers who are <strong>strictly </strong><strong>more than 60 years old</strong>.</em></p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> details = [&quot;7868190130M7522&quot;,&quot;5303914400F9211&quot;,&quot;9273338290F4010&quot;]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong> The passengers at indices 0, 1, and 2 have ages 75, 92, and 40. Thus, there are 2 people who are over 60 years old.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> details = [&quot;1313579440F2036&quot;,&quot;2921522980M5644&quot;]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> None of the passengers are older than 60.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= details.length &lt;= 100</code></li>
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<li><code>details[i].length == 15</code></li>
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<li><code>details[i] consists of digits from &#39;0&#39; to &#39;9&#39;.</code></li>
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<li><code>details[i][10] is either &#39;M&#39; or &#39;F&#39; or &#39;O&#39;.</code></li>
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<li>The phone numbers and seat numbers of the passengers are distinct.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def countSeniors(self, details: List[str]) -> int:
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return sum(int(x[11:13]) > 60 for x in details)
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```
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### **Java**
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```java
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class Solution {
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public int countSeniors(String[] details) {
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int ans = 0;
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for (var x : details) {
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int age = Integer.parseInt(x.substring(11, 13));
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if (age > 60) {
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++ans;
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countSeniors(vector<string>& details) {
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int ans = 0;
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for (auto& x : details) {
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int age = stoi(x.substr(11, 2));
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ans += age > 60;
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func countSeniors(details []string) (ans int) {
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for _, x := range details {
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age, _ := strconv.Atoi(x[11:13])
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if age > 60 {
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ans++
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}
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}
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return
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2600-2699/2678.Number of Senior Citizens/Solution.cpp

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class Solution {
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public:
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int countSeniors(vector<string>& details) {
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int ans = 0;
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for (auto& x : details) {
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int age = stoi(x.substr(11, 2));
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ans += age > 60;
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}
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return ans;
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}
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};

‎solution/2600-2699/2678.Number of Senior Citizens/Solution.go

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func countSeniors(details []string) (ans int) {
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for _, x := range details {
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age, _ := strconv.Atoi(x[11:13])
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if age > 60 {
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ans++
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}
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}
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return
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}

‎solution/2600-2699/2678.Number of Senior Citizens/Solution.java

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class Solution {
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public int countSeniors(String[] details) {
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int ans = 0;
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for (var x : details) {
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int age = Integer.parseInt(x.substring(11, 13));
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if (age > 60) {
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++ans;
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}
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}
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return ans;
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}
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}

‎solution/2600-2699/2678.Number of Senior Citizens/Solution.py

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class Solution:
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def countSeniors(self, details: List[str]) -> int:
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return sum(int(x[11:13]) > 60 for x in details)

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