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
想讓結果最小很容易想到貪心算法,將各串數字按序排列輸出即可。做題時想的比較複雜,兩串字符串公共長度處依次比較字典序後,再將剩餘字符串不斷取餘比較,雖然結果正確,但浪費不少時間。題解的方法非常簡便,對數字串S1、S2,若S1+S2<S2+S1,則S1排在S2前面,否則排在S2後面。
題解方法:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Comparator;
class Cmp implements Comparator<String>{
@Override
public int compare(String o1, String o2) {
return (o1+o2).compareTo(o2+o1);
}
}
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader reader=new BufferedReader(new InputStreamReader(System.in));
String[] line=reader.readLine().split(" ");
int N=Integer.valueOf(line[0]);
Arrays.sort(line,1,line.length,new Cmp());
StringBuilder sb=new StringBuilder();
for(int i=1;i<line.length;i++) {
sb.append(line[i]);
}
String ans=sb.toString();
//找出第一個不爲0的數
int begin=0;
while(begin<ans.length()) {
if(ans.charAt(begin)!='0')
break;
begin++;
}
if(begin==ans.length())
System.out.println(0);
else {
System.out.println(ans.substring(begin));
}
}
}
我的比較函數:
class Cmp implements Comparator<String>{
@Override
public int compare(String o1, String o2) {
int len=Math.min(o1.length(),o2.length());
int i=0;
while(i<len) {
//公共長度部分升序排序
if(o1.charAt(i)!=o2.charAt(i))
return o1.charAt(i)-o2.charAt(i);
i++;
}
//取餘比較
if(o1.length()==o2.length()) {
return 0;
}else if(o1.length()>o2.length()) {
return compare(o1.substring(i), o2);
}else {
return compare(o1, o2.substring(i));
}
}
}