如何为具有交换功能的双向链表创建冒泡排序？

Question

我试图弄清楚如何为 DNode 编写一个名为 DNode *DNode::bubble() 的成员函数 DNode *head 可以对链表进行冒泡排序并返回新的头部。 我想使用成员函数void swap(DNode *that) 交换两个节点。

以下是我目前对 .h 和 .cpp 文件的了解：

#ifndef DNODE_H
#define DNODE_H
class DNode {

public:

int key;
DNode *next;
DNode *prev;

void add(int key);
void print();
DNode();
DNode *bubble();
void swap(Dnode *that);
};

#endif

#include "DNode.h"
#include <cstdlib>
#include <iostream>
using namespace std;

void DNode::swap(DNode *that){
DNode *temp = this;
DNode *R1, *L1, *R2, *L2;

}

Dnode *Dnode::bubble(){
DNode *temp = this;
int count = 0;
while(start != NULL){
count++;
start = start->next;
}

for(int i = 0; i < count; i++){

}
}

我遇到的问题是交换函数，它只用那个参数交换列表中的节点。

Answer 1

您需要做的就是调整两者的next 和prev 成员，并同步它们的上一个和下一个元素的链接：

void DNode::swap(DNode *that) 
{
  // this->prev->next points back to 'this' should become 'that'
  if(this->prev) {
    this->prev->next = that;
  }
  // this->next->prev points back to 'this' should become 'that'
  if(this->next) {
    this->next->prev = that;
  }
  // that->prev->next points back to 'that' should become 'this'
  if(that->prev) {
     that->prev->next = this;
  }
  // that->next->prev points back to 'that' should become 'this'
  if(that->next) {
     that->next->prev = this; 
  }
  // remember whatever 'this' ->next and ->prev point to
  DNode * n1 = this->next, * p1 = this->prev;
  // let 'this' take the position of 'that in the list
  this->prev = that->prev;
  this->next = that->next;
  // let 'that' take the position of the original 'this' in the list.
  that->prev = p1;
  that->next = n1;
}

或者：如果您只是想在列表中的特定逻辑位置交换 值，那么您也可以简单地交换值：

void swap(DNode* that)
{
  int old = this->key;
  this->key = that->key;
  that->key = old;
}