將鏈接列表按字母順序排序

Question

從排序的鏈接列表中刪除節點時遇到問題。我從.txt文件讀入了73個必須按字母順序排序的不同名稱。我有一個switch語句應該能夠對鏈表執行5個單獨的事情。目前我已經拿到1號和2號的工作，但不是三號。 ＃3希望我能夠從鏈接列表中刪除名稱。在我輸入要刪除的名稱後，我的代碼不會顯示任何內容。因此我假設我遇到了deleteAfter函數的問題。任何人都可以給我一個暗示，爲什麼這可能是？

#include "stdafx.h" 
#include <iostream> 
#include <string> 
#include <fstream> 
using namespace std; 

struct node{ 
    string name; 
    node *next; 
}; 

node *A = NULL; 

void addnode(string newname){ 
    node *add, 
     *last, 
     *current; 

    add = new node; 
    add->name = newname; 

    if (A == NULL){ 
     add->next = A; 
     A = add; 
    }else{ 
     current = A; 
     last = A; 
     while (current && current->name < newname){ 
      last = current; 
      current = current->next; 
     } 

     if (current == A){ 
      /* Insert before 1st node */ 
      add->next = A; 
      A = add; 
     } 
     else{ 
      /* Insert between last and current 
       or at the end of the list */ 
      last->next = add; 
      add->next = current; 
     } 
    } 
} 
void deleteName(string name) 
{ 
    node *curr; 
    node *nextNode; 
    curr = A; 
    nextNode = curr; 
    while(curr){ 
     if(curr -> next -> name == name){ 
      nextNode = curr -> next; 
      curr -> next = nextNode -> next; 
     } 

    } 


} 



void display() 
{ 
    node *curr; 
    curr = A; 
    while(curr){ 
     if(A == NULL){break;} 
     cout << A->name << endl; 
     A = A->next; 
    } 

} 

int main(){ 


    int input, count; 
    count = 0; 
    ifstream dataFile; 
    dataFile.open("Data.txt"); 
    string item; 
    string name; 
    while(dataFile) 
    { 
     dataFile >> item; 
     addnode(item); 
     count++; 
    } 




    cout << "1. Display the linked list\n"; 
    cout << "2. Display the length of the list\n"; 
    cout << "3. Delete name from the list\n"; 
    cout << "4. display the length of a section of the list\n"; 
    cout << "5. Print out section of list\n"; 
    cin >> input; 

    switch (input) 
    { 
    case 1: 
     display(); 
     break; 
    case 2: 
     cout << "There are " << count - 1 << " names in the list\n"; 
     break; 
    case 3: 
     cout << "Type in the name that you want to be deleted: "; 
     cin >> name; 
     deleteName(name); 
     display(); 
     break; 
    case 4: 
     break; 
    case 5: 
     break; 
    } 


    system("PAUSE"); 
    return 0; 

} 

這是我到目前爲止的代碼。你會注意到，在我的主函數中，我從一個名爲「Data.txt」的文件讀取輸入。

joe 
bob 
harry 
mary 
brian 
tom 
jerry 
bullwinkle 
pam 
ellis 
dale 
bill 
barrack 
george 
gertrude 
zack 
zeus 
apollo 
gemini 
greg 
larry 
meriam 
webster 
thomas 
stewart 
dianna 
theresa 
billyjoe 
carl 
karl 
charles 
karla 
donna 
tena 
kerry 
howard 
johnson 
ulyssess 
paul 
peter 
issaac 
marvin 
dudz 
chuck 
ellie 
anny 
judy 
matt 
ross 
dan 
robert 
kim 
eric 
junkun 
ghassan 
cris 
raymond 
avery 
roy 
halley 
mitzee 
ziggy 
rocky 
twirly 
max 
huey 
dewy 
hongkongfooey 
clarence 
lala 
sammy 
fred 
francis 

這就是txt文件由^^組成的內容。任何建議將不勝感激。謝謝！

Answer 1

您正在訪問下一個，但未檢查它是否爲空，而且您沒有遍歷該列表。另外，你應該在找到它之後再打破它（除非你想刪除所有的實例，並且你應該刪除這個節點，因爲你會泄漏內存，而且你也不能刪除第一個元素，因爲你永遠不會檢查它。因爲你需要處理不斷變化的根節點你可以在它的特殊檢查添加。

if (A != nullptr && A->name == name) 
{ 
    node *toBeDeleted = A; 
    A = A->next; 
    delete toBeDeleted; 
    return; 
} 

while(curr && curr->next){ 
    if(curr->next->name == name){ 
     nextNode = curr->next; 
     curr->next = nextNode->next; 
     delete nextNode; 
     break; 
    } 
    curr = curr->next; 
} 

當然，如果你想刪除的名稱的所有實例，您需要刪除的回報和休息聲明

你的顯示功能也將清空列表您需要設置CURR，沒有A：

void display() 
{ 
    node *curr; 
    curr = A; 
    while(curr){ 
     cout << curr->name << endl; 
     curr = curr->next; 
    } 
} 

Answer 2

while (current && strcmp(current->name , newname) <=0){ 
    last = current; 
    current = current->next; 
} 

試試這個。

Answer 3

您正在使用鏈接列表數據結構。我發現奇怪的是你使用了一個循環。最後一個節點的下一個元素再次指向開頭。

這裏是deleteName我根據自己的知識和風格（即我相信看到）的級建議：

void deleteName(string name) { 

    node *current = A; 
    node *previous; 

    while (current) { 
     if (current->name == name) { 
      previous->next = current->next; 
      delete current; 
      break; 
     } else { 
      previous = current; 
      current = current->next; 
     } 
    } 
}