如何解决比较链接列表，C ++，相同顺序但可以不同
|| 假设我们得到了函数定义 bool sameValsOrder（节点* p，节点* q） 我们必须编写一个函数来比较两个链表，如果它们具有相同的顺序，则返回true，否则返回false [77777]和[77]->是 [1234567]和[1234555567]-> truebool sameValsOrder (node *p,node *q)
{
if ( q==NULL && p==NULL)
return true;
else if ((q=NULL && p != NULL)|| (p=NULL && q!= NULL))
return false;
else if ( q != NULL && p != NULL)
while ( q != 0 && p != 0)
{
if (p->data != q->data)
{
return false;
break;
}
else
{
p= p-> next ;
q= q-> next;
}
}
return true;
}
解决方法
根据您所写的内容，您实际上并不关心这些值，但是如果要订购列表，您想返回true吗？似乎您只需要遍历列表中的每个值即可。只要NEXT值不小于PREVIOUS值，就继续遍历该列表。如果到达末尾，则返回true，因为列表是按顺序排列的。如果在任何时候遇到的值都小于以前的任何值，则只需在此处返回false即可。
#include <iostream>
using namespace std;
class node{
public:
node(){next = NULL;}
int data;
node * next;
};
class myList{
public:
myList(){pRoot = NULL;}
void add(int data);
node * pRoot;
};
bool sameValsOrder (node *p,node *q)
{
if ( q==NULL && p==NULL) // If both lists are empty
return true;
else if ((q==NULL && p != NULL)|| (p==NULL && q!= NULL)) // One list is empty and the other is not
return false;
else if ( q != NULL && p != NULL) //Both lists contain values we must check
{
int temp; //I am going to assume a singly linked list (no access to previous value),need this to hold previous value
temp = p->data;
while (p->next != NULL) //The list still contains elements
{
if (p->next->data < temp) //The value in the current node is LESS than our temp,then list is out of order so return false
return false;
else { //Otherwise move to the next node
temp = p->data;
p = p->next;
}
}
temp = q->data; //Reset temp for q
//Do the same thing for q
while (q->next != NULL) //The list still contains elements
{
if (q->next->data < temp) //The value in the current node is LESS than our temp,then list is out of order so return false
return false;
else { //Otherwise move to the next node
temp = q->data;
q = q->next;
}
}
}
return true; //If we are this are then both lists should be ordered
}
int main()
{
myList * p = new myList();
myList * q = new myList();
p->add(7);
p->add(6);
p->add(5);
p->add(4);
p->add(3);
p->add(2);
p->add(1);
q->add(7);
q->add(6);
q->add(5);
q->add(5);
q->add(5);
q->add(5);
q->add(4);
q->add(3);
q->add(2);
q->add(1);
cout << sameValsOrder (p->pRoot,q->pRoot) << endl;
return 0;
}
void myList::add(int data)
{
node * nodeToAdd = new node();
nodeToAdd->data = data;
if(pRoot == NULL) //List is empty
{
pRoot = nodeToAdd;
pRoot->next = NULL;
}
else //List not empty insert new node at beginning
{
nodeToAdd->next = pRoot;
pRoot = nodeToAdd;
}
}
while ( q != 0 && p != 0)
{
if (p->data != q->data)
return false;
break;
else
p= p-> next ;
q= q-> next;
}
while(q!=NULL || p!=NULL)
{
if(q->data==p->data)
{
p=p->next;
break;
}
else(q->data < p->data)
q=q->next;
}
bool sameValsOrder (node *p,node *q)
{
while not at the end of either list,and the current nodes are the same {
skip run in q,taking care to deal with the NULL at the end of the list
skip run in p,taking care to deal with the NULL at the end of the list
}
if we\'ve reached the end of both lists,they are equivalent
}
bool sameValsOrder (node *p,node *q)
{
while (q && p && (q->data == p->data)) {
/* find next different node in each list (ie.,skip runs) */
int tmp = q->data; /* or whatever type `data` is */
while (q && (q->data == tmp)) {
q = q->next;
}
while (p && (p->data == tmp)) {
p = p->next;
}
}
/*
* we\'ve either reached the end of one or both lists,* or have found a `data` difference
*/
if (p == q) {
/* should happen only when `q` and `p` are NULL */
assert(q == NULL);
assert(p == NULL);
/* we\'ve reached the end of both lists,so they\'re \'equivalent\' */
return true;
}
return false;
}
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