[Leetcode] 2678.Number of Senior Citizens

題目

You are given a 0-indexed array of strings details. Each element of details provides information about a given passenger compressed into a string of length 15. The system is such that:

• The first ten characters consist of the phone number of passengers.
• The next character denotes the gender of the person.
• The following two characters are used to indicate the age of the person.
• The last two characters determine the seat allotted to that person.

Return the number of passengers who are strictly more than 60 years old.

2678.Number of Senior Citizens

範例

Example 1:

Input: details = [“7868190130M7522”,”5303914400F9211”,”9273338290F4010”]
Output: 2
Explanation: 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.

Example 2:

Input: details = [“1313579440F2036”,”2921522980M5644”]
Output: 0
Explanation: None of the passengers are older than 60.

解題思路

陣列裡的每一個字串長度都為15，前10個字元為電話，第11個字元為性別(M、F、0)，第12、13字元為年齡，最後2個為座位，可以確定都是固定長度。根據題目要求，只需要判斷年齡是否大於60，所以只要取出第12、13個字元做處理就好，抓取字串內的某幾個字元，可以使用C++內建函式substr或是一般陣列。參考 C++ substr() 使用方式

題解

1. 用 for 迴圈走訪陣列裡的字串
2. 可以使用 substr 取出子字串或是使用陣列方式
3. 將字串轉換為整數 int
4. 判斷是否大於 60 ，並計數
5. 回傳大於 60 總數量

方法一 :

class Solution {
public:
    int countSeniors(vector<string>& details) {
        
        int count = 0;
        for(const string &s : details) {
            string sub_str = s.substr(11, 2);
            if(stoi(sub_str) > 60)
                count++;
        }
        return count;
    }
};

方法二(不使用substr) :

class Solution {
public:
    int countSeniors(vector<string>& details) {
        
        int res = 0;
        for(int i = 0; i < details.size(); i++){
            int age = ((details[i][11] - '0') * 10 + details[i][12] - '0');
            res += (age > 60);
        }
        return res;
    }
};