PTA甲級 1038 Recover the Smallest Number (C++)

Given a collection of number segments, you are supposed to recover the smallest number from them. For example, given { 32, 321, 3214, 0229, 87 }, we can recover many numbers such like 32-321-3214-0229-87 or 0229-32-87-321-3214 with respect to different orders of combinations of these segments, and the smallest number is 0229-321-3214-32-87.

Input Specification:

Each input file contains one test case. Each case gives a positive integer N ( ≤ 1 0 4 ) N (≤10^4) N(≤104) followed by N N N number segments. Each segment contains a non-negative integer of no more than 8 digits. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the smallest number in one line. Notice that the first digit must not be zero.

Sample Input:

5 32 321 3214 0229 87
           

Sample Output:

22932132143287
           

Caution:

這道題想了好久都沒有比較好的思路，遂上網查了查，看到這個題解，真的一語驚醒夢中人。

下面是搬運的這句話：

直接字元串排序然後合并再去除前導0，但排序規則不是直接比較大小，而是若S1+S2<S2+S1則S1排在S2前

Solution:

// Talk is cheap, show me the code
// Created by Misdirection 2021-08-16 18:01:28
// All rights reserved.

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>

using namespace std;

bool cmp(string& s1, string& s2){
    return s1 + s2 < s2 + s1;
}

int main(){
    int n;
    scanf("%d", &n);
    string *nums = new string[n];

    for(int i = 0; i < n; ++i) cin >> nums[i];

    sort(nums, nums + n, cmp);

    string ans = "";
    for(int i = 0; i < n; ++i) ans += nums[i];
    
    while(ans[0] == '0') ans = ans.substr(1, ans.length() - 1);

    if(ans.length() == 0) ans = "0";
    printf("%s\n", ans.c_str());

    return 0;
}