Build the next version of the Unix sort command, this time with the -n flag baked in. Each input line is the string form of an integer that fits in 32 bits (an optional leading -, no leading +; leading zeros such as "007" are allowed). Sort the lines numerically (not lexicographically) in ascending order, and return the original strings joined by newlines (\n). Lines with the same numeric value keep their original relative order (a stable sort). The real sort -n would break such ties by comparing the text; this version does not. If the list is empty, return the empty string.
Note: lexicographic sort would put "100" before "20"; numeric sort puts 20 before 100.
Function signature:
string sortLines(vector<string> lines)
Example:
Input: lines = ["100", "20", "3", "10"] Output: "3\n10\n20\n100" When printed: 3 10 20 100
Pass a lambda comparator to
std::stable_sort (not plain std::sort, which is not stable) that converts each operand to int via stoi and compares numerically.
The full solution:
class Solution {
public:
string sortLines(vector<string> lines) {
vector<string> sortedLines = lines;
stable_sort(sortedLines.begin(), sortedLines.end(), [](const string& a, const string& b) {
return stoi(a) < stoi(b);
});
string result;
for (int i = 0; i < sortedLines.size(); i++) {
if (i > 0) result += "\n";
result += sortedLines[i];
}
return result;
}
};
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