問題
Data typed by people is messy. You get a list of strings and must add up the ones that are whole numbers, ignoring the rest instead of crashing.
An entry counts if, after removing spaces at both ends, it is an optional + or - sign followed by one or more digits. Anything else (empty text, decimals such as "2.5", words) is skipped.
Implement sum_valid_numbers(values).
Input
values: The entries to read.
Output
Return the sum of the valid whole numbers (0 if there are none).
制約
0 ≤ number of entries ≤ 1,000
Each entry has at most 20 characters; valid numbers fit in a 64-bit integer
例
入力：
values = ["10", " 5 ", "abc", "-3"]
期待される出力：
expected = 12入力：
std::vector<std::string> values = {"10", " 5 ", "abc", "-3"};
期待される出力：
long long expected = 12;入力：
string[] values = new string[] {"10", " 5 ", "abc", "-3"};
期待される出力：
long expected = 12;解説
"10", " 5 " (spaces are trimmed) and "-3" are whole numbers; "abc" is skipped. 10 + 5 - 3 = 12.
入力：
values = ["2.5", "", "seven"]
期待される出力：
expected = 0入力：
std::vector<std::string> values = {"2.5", "", "seven"};
期待される出力：
long long expected = 0;入力：
string[] values = new string[] {"2.5", "", "seven"};
期待される出力：
long expected = 0;解説
None of the entries is a whole number, so the sum is 0.
解き方のヒント
まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。
ヒント 1
Strip the spaces first, then decide whether what is left is a valid whole number.
ヒント 2
Converting inside try/except (or checking the characters) lets you skip bad entries without stopping.
ヒント 3
Remember that a lone sign such as "+" is not a number.
目安の計算量
O(total length) time, O(1) extra space
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。