DS and Algorithm_session_03_doubly lingked list.ppt

* * * * * * * * * * Tell the students that a circular linked list can be a singly circular or a doubly circular linked list. * * * * Ask the students to write the algorithm first and then come up with the solution. * * * * * * * Tell the students that inserting a node in the middle of a circular linked list is same as that of a singly-linked list. This is because the only difference between the two lists is with the arrangement in the address field of the last node. Therefore, the process of insertion of a node at the end of the two lists is different. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * In this slide you need to show the calculation to determine the sum of an arithmetic progression for bubble sort algorithm. Refer to student guide. * * * * * * * * * * * * * * * -8 8 theList 3 14 0 -1 q p(x)=3x14-8x8+6x2+2 输入多项式的项，建立多项式: 6 2 (3,14) 2 0 (-8,8) (6,2) (2,0) Output函数将多项式按降幂方式送输出流。 void Polynominal::Output(ostream out)const { int first=1; Term *p=theList-link; coutThe polynominal is:\nendl; for ( ; p!=theList; p=p-link) { if (!first (p-coef0)) out+; first=0; out*p; // 调用Term类上重载的“”操作。 } cout\nendl; } 3 14 px -8 8 2 0 6 2 0 -1 3. 多项式相加——实现q(x)?q(x)+p(x) 设p和q分别指向多项式p(x)和q(x)的当前正进行比较的项，初始 时分别指向两多项式中最高幂次的项。q1指向q的前驱结点。 对p(x)进行遍历，根据指针p、q的exp域的大小情况做相应处理： ①p-expq-exp 则q指示的项应成为结果多项式中的一项，所以q1和q右移一项。 ②p-exp==q-exp 则系数相加，即q-coef=q-coef+p-coef。如果q-coef不为零，则指 针q1和q 均右移一个结点；否则从q(x)中删除q指示的结点。 ③p-expq-exp 则复制p所指示的结点，并将其插在 q1之后。 指针p右移一项； 重复上述处理，直到p(x)中全部结点都处理完结束。 p q1 3 14 px -8 8 2 0 6 2 0 -1 q 6 12 qx 8 8 3 2 -2 6 0 -1 3 14 qx 2 10 2 0 -4 8 0 -1 一元整系数多项式相加 void Polynominal::PolyAdd(Polynominal r) //将多项式r加到多项式this上 { Term* q,*q1=theList,*p; //q1指向表头结点 p=r.theList-link; //p指向第一个要处理的结点 q=q1-link; //q1是q的前驱，p和q就指向两个当前进行比较的项 while (p-exp=0) //对r的单循环链表遍历,直到全部结点都处理完 { while (p-expq