1038. Recover the Smallest Number (30)
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 (<=10000) 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. Do not output leading zeros.
Sample Input:5 32 321 3214 0229 87Sample Output:
22932132143287
很明顯這是一道排序題,這道題的學問主要在排序函數,不能使用通常的字符串比較函數,例如實例中,當32,321,3214這3個數同時存在於序列中時,如何決定這三個數的順序,可以肯定的是,當多個數有相同的前綴時,這幾個數出現時一定是相鄰的,如結果中的321 32 3214,方法就是當比較兩個字符串大小時如32和321,如果有相同前綴,則刪除較長的字符串中的前綴,並比較剩下的數的大小,即32和321的比較結果應該是32和1的比較結果,同理,321和3214的比較結果應該是321和4的比較結果,3214和32的比較結果應該等於14和32的比較結果,代碼如下:
本題目的另外一個坑點就是當所有的數都是0的時候,去除前導的0可能會使得整個數都被剪裁沒了,此時要進行特判,否則可能會發生“異常退出”的情況
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
using namespace std;
vector<string> num;
bool cmp(string a, string b){
int len[2];
len[0] = a.length(), len[1] = b.length();
int i = 0;
while (i < len[0] && i < len[1]){
if (a[i] < b[i])
return true;
else if (a[i] > b[i])
return false;
else;
i++;
}
if (i < len[0])
return cmp(a.substr(i),b);
else if (i < len[1])
return cmp(a,b.substr(i));
else
return true;//兩個數完全相同
}
int main(){
string temp, ans;
int n,k;
cin >> n;
vector<string>::iterator it;
while (n--){
cin >> temp;
num.push_back(temp);
}
sort(num.begin(), num.end(), cmp);
it = num.begin();
while (it != num.end()){
ans += *it;
it++;
}
if ((k = ans.find_first_not_of('0')) != string::npos)//如果整個數都是0
ans = ans.substr(k);
else
ans = "0";
cout << ans << endl;
return 0;
}