Same signature as v2/v3, but now n may be negative. Real head -n -K prints all lines except the last K. So when n < 0 return all lines except the last |n|; when n ≥ 0 behave exactly like v3 (return the first n, capped at the input length). If |n| is at least the number of lines, nothing is left, so return the empty string.
Function signature:
string head(vector<string> lines, int n)
Example (negative n):
Input: lines = ["a", "b", "c", "d", "e"], n = -2 Output: "a\nb\nc" When printed: a b c
Branch on the sign of
n: when n < 0, the end index is (int)lines.size() + n (with a max(0, ...) floor); otherwise it's min((int)lines.size(), n).
int end;
if (n < 0) {
end = max(0, (int)lines.size() + n);
} else {
end = min((int)lines.size(), n);
}
The full solution:
class Solution {
public:
string head(vector<string> lines, int n) {
int end;
if (n < 0) {
end = max(0, (int)lines.size() + n);
} else {
end = min((int)lines.size(), n);
}
string result;
for (int i = 0; i < end; i++) {
if (i > 0) result += "\n";
result += lines[i];
}
return result;
}
};
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