Java程序练习-长整数加法运算

长整数加法运算
时间限制: 2000ms内存限制: 2000kB
描述
问题描述：设计一个实现任意长的整数进行加法运算的演示程序。

基本要求：利用双向循环链表实现长整数的存储，每个结点含一个整形变量。任何整形变量的范围是 -(2^15 - 1)~(2^15 - 1)。输入和输出形式：按中国对于长整数的表示习惯，每四位一组，组间用逗号隔开。
输入
-9999,9999（输入第一个加数）
1,0000,0000,0000（输入第二个加数）
输出
9999,0000,0001（输出结果,如果前面有多余的0应该去掉）
样例输入
-9999,9999 1,0000,0000,0000
样例输出
9999,0000,0001
参考代码

//import java.io.BufferedReader; import java.io.IOException; //import java.io.InputStreamReader; import java.util.ArrayList; import java.util.Iterator; import java.util.List; import java.util.Scanner; public class Main { public static void main(String[] args) throws IOException { //BufferedReader cin = new BufferedReader(new InputStreamReader(System.in)); Scanner cin = new Scanner(System.in); while(cin.hasNext()){ //String s1 = cin.readLine(); //String s2 = cin.readLine(); String s1 = cin.next(); String s2 = cin.next(); boolean minus_1 = false; boolean minus_2 = false; if(s1.contains("-")) minus_1 = true; if(s2.contains("-")) minus_2 = true; if(minus_1 && !minus_2){ calculate(s1,s2,'-',1); }else if(!minus_1 && minus_2){ calculate(s2,s1,'-',1); }else if(minus_1 && minus_2){ calculate(s1,s2,'+',2); }else{ calculate(s1,s2,'+',3); } } } private static void calculate(String s1, String s2, char c, int kind) { String s1a[] = s1.split(","); String s2a[] = s2.split(","); List<Integer>list1 = new ArrayList<Integer>(); List<Integer>list2 = new ArrayList<Integer>(); List<Integer>list_temp = new ArrayList<Integer>(); List<Integer>list3 = new ArrayList<Integer>(); boolean flag = false; switch(kind){ case 1: s1a[0] = String.valueOf(0 - Integer.parseInt(s1a[0])); break; case 2: s1a[0] = String.valueOf(0 - Integer.parseInt(s1a[0])); s2a[0] = String.valueOf(0 - Integer.parseInt(s2a[0])); break; case 3: break; } int k = 0; boolean pout = false; for(int i = 0;i <s1a.length;++ i){ int n = Integer.parseInt(s1a[i]); if(n > 0 || pout){ pout = true; list_temp.add((n + k) % 10000); } k = n / 10000; } for(int i = list_temp.size() - 1;i >= 0;-- i){ list1.add(list_temp.get(i)); } k = 0; pout = false; list_temp.clear(); for(int i = 0;i < s2a.length;++ i){ int n = Integer.parseInt(s2a[i]); if(n > 0 || pout){ list_temp.add((n + k) % 10000); pout = true; } k = n / 10000; } for(int i = list_temp.size() - 1;i >= 0;-- i){ list2.add(list_temp.get(i)); } Iterator<Integer> it1,it2; if(list1.size() == 0){ list1.add(0); } if(list2.size() == 0){ list2.add(0); } int n = 0; //System.out.println(list1); //System.out.println(list2); switch(c){ case '-': if(list1.size() < list2.size()){ list_temp = list1; list1 = list2; list2 = list_temp; flag = true; }else if(list1.size() == list2.size()){ for(int i = list1.size() - 1;i >= 0;-- i){ if(list1.get(i) > list2.get(i)){ flag = true; break; } } if(!flag){ list_temp = list1; list1 = list2; list2 = list_temp; flag = true; }else{ flag = false; } } n = 0; int j = 0; int m = 0; it1 = list1.iterator(); it2 = list2.iterator(); //System.out.println(list1); //System.out.println(list2); while(it1.hasNext()||it2.hasNext()){ if(it1.hasNext()){ n = it1.next() + m; j ++; } if(it2.hasNext()){ int temp = it2.next(); if(n >= temp){ n -= temp; m = 0; }else{ n = n + 10000 - temp; m = -1; } } if(n < 0){ n += 10000; m = -1; } if(n != 0 || j != list1.size()) list3.add(n); } break; case '+': n = 0; it1 = list1.iterator(); it2 = list2.iterator(); while(it1.hasNext()||it2.hasNext()){ if(it1.hasNext()){ n += it1.next(); } if(it2.hasNext()){ n += it2.next(); } list3.add(n % 10000); n /= 10000; } if(n != 0){ list3.add(n); } break; default: break; } int p = 0; //System.out.println(list3); if(list3.size() == 0){ list3.add(0); } for(int i = list3.size() - 1;i >= 0;-- i) if(list3.get(i) != 0){ p = i; break; } n = list3.get(p); switch(kind){ case 1: //System.out.println(flag); if(!flag){ System.out.printf("%d",-n); }else{ System.out.printf("%d",n); } break; case 2: System.out.print("-"); System.out.printf("%d",list3.get(p)); break; case 3: System.out.printf("%d",list3.get(p)); break; } for(int i = p - 1;i >= 0;-- i) System.out.printf(" %04d",list3.get(i)); System.out.println(); } }