【PAT甲级】Recover the Smallest Number

2019 年 11 月 8 日
筆記

本文链接：https://blog.csdn.net/weixin_42449444/article/details/89716717

Problem Description：

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 (≤104) followed by 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

解题思路：

建立一个vector用来存放string型的数字，自定义一个比较规则cmp来使数字构成的字符串经可能小，然后以cmp为比较规则来对vector进行sort。接着判断字符串是不是以0开头，若字符串以0开头则需要删除0。然后判断字符串是不是为空（即字符串长度是否为0），若字符串str为空，则令str = "0"，最后输出str即可。

AC代码：

#include <bits/stdc++.h>  using namespace std;    bool cmp(string s1,string s2)   //使数字构成的字符串要尽可能小  {      return s1+s2 < s2+s1;  }    int main()  {      int N;      cin >> N;      vector<string> v;      for(int i = 0; i < N; i++)      {          string temp;          cin >> temp;          v.push_back(temp);      }      sort(v.begin(), v.end(),cmp);      string str = "";      for(auto it : v)      {          str += it;      }      int len = str.length();      while(len !=0 & str[0] == '0')   //一大串数字组成的字符串第一个数字不能为0      {          str.erase(str.begin());      }      if(len == 0)      {          str = "0";      }      cout << str << endl;      return 0;  }