Java数据结构的题目
求多项式3x4-6x2+5x-10加上多项式-3x5+7x4+x3+6x2的结果

求代码 谢谢
[解决办法]
思路：
建立一个类，具有系数和次数两种属性；先把次数作为关键字，使得两个多项式排序成为两个有序数组；接下来就是归并有序数组的问题了。
代码你自己实现吧，这是掌握顺序表的必会题，也是学习二路归并排序的基础。
(如果还想锻炼一下的话，自己用正则表达式实现“自动取得多项式的所有的系数和次数”)

[解决办法]

public class PostFix {

    public static String postFix(String expression) {
CS401StackLinkedListImpl<Character> operandStack = new CS401StackLinkedListImpl<Character>();
String postfix = "";
for (int i = 0; i < expression.length(); i++) {
    char c = expression.charAt(i);
    Character temp;
    switch (c) {
    case ' ':
break;

    case '+':
    case '-':
while (operandStack.size() != 0) {
    temp = operandStack.pop();
    if (temp == '(') {
operandStack.push(temp);
break;
    }
    postfix += " " + temp;
}
operandStack.push(c);
postfix += " ";
break;

    case '*':
    case '/':
while (operandStack.size() != 0) {
    temp = operandStack.pop();
    if (temp == '(' 
[解决办法]
 temp == '+' 
[解决办法]
 temp == '-') {
operandStack.push(temp);
break;
    }
    postfix += " " + temp;
}
operandStack.push(c);
postfix += " ";
break;

    case '(':
operandStack.push(c);
break;

    case ')':
while (operandStack.size() != 0) {
    temp = operandStack.pop();
    if (temp == '(') {
break;
    }
    postfix += " " + temp;
}
break;

    default:
postfix += c;
break;
    }
}

while (operandStack.size() != 0) {
    postfix += " " + operandStack.pop();
}
return postfix;
    }

    public static int calculateArithmeticExp(String postfix) {
CS401StackLinkedListImpl<Integer> stack = new CS401StackLinkedListImpl<Integer>();
String[] strings = postfix.split(" ");

for (int i = 0; i < strings.length; i++) {
    String temp = strings[i].trim();
    if (temp == "")
continue; 
 
    else if (temp.matches("\\d+")) {
stack.push(Integer.parseInt(temp));
    } else {
int x = stack.pop();
int y = stack.pop();
if (temp.equals("+"))
    stack.push(x + y);
else if (temp.equals("-"))
    stack.push(y - x);
else if (temp.equals("*"))
    stack.push(x * y);
else if (temp.equals("/"))
    stack.push(y / x);
    }

}
return stack.pop();

    }

}

我写过的作业，自己改下linkedlist 和 stack 之类的，记得导包
[解决办法]

public static int countString(String s){
System.out.println(s);
int num = 0 ; 
String[] ss =s.split("[\\+\\-\\*/\\(\\)]") ;
if( ss.length>2){
int si = 0 ;
if( (si=s.indexOf("("))!=-1){
num =countString(s.substring(si+1 , s.indexOf(")")));
num =countString(s.substring(0 , si)+num + s.substring(s.indexOf(")")+1));
}else if((si=s.replaceAll("[\\*/]", ",").indexOf(","))!= -1 ){
int statr = s.substring(0, si).replaceAll("[\\+\\-\\*/\\(\\)]", ",").lastIndexOf(",") ;
System.out.println(si);
int end = s.substring( si+1).replaceAll("[\\+\\-\\*/\\(\\)]", ",").indexOf(",") ;
num =countString(s.substring(statr==-1?0:statr+1, end==-1?s.length():si+end+1));
num =countString(s.substring(0 , statr==-1?0:statr+1)+num + s.substring(end==-1?s.length():si+end+1));
}else if((si=s.replaceAll("[\\+\\-]", ",").indexOf(","))!= -1 ){
if(si==0){
num =  countString("&"+s.substring(1) );
}else{
int statr = s.substring(0, si).replaceAll("[\\+\\-\\*/\\(\\)]", ",").lastIndexOf(",") ;
int end = s.substring( si+1).replaceAll("[\\+\\-\\*/\\(\\)]", ",").indexOf(",") ;
num =countString(s.substring(statr==-1?0:statr+1, end==-1?s.length():end+2));
num =  countString(s.substring(0 , statr==-1?0:statr+1)+num + s.substring(end==-1?s.length():end+2));
}
}

}else{
int index =s.replaceAll("[\\+\\-\\*/\\(\\)]", ",").indexOf(",") ;
if(index!=-1){
String s2 =s.substring(index,index+1) ;
if(ss[0].substring(0, 1).equals("&"))
ss[0] ="-"+ ss[0].substring(1) ;
if(s2.equals("*"))
num = new Integer(ss[0])  * new Integer(ss[1]) ;
if(s2.equals("/"))
num = new Integer(ss[0])  / new Integer(ss[1]) ;
if(s2.equals("+"))
num = new Integer(ss[0])  + new Integer(ss[1]) ;
if(s2.equals("-"))
num = new Integer(ss[0]) - new Integer(ss[1]) ;
}else{
num = new Integer(s)  ;
}
}

return num;
}

试试

[解决办法]
http://blog.csdn.net/ss36290109/article/details/8235309 有用么。