鏈表面試題2

有關單鏈表的面試題2

複雜鍊錶的複製

//複雜鍊錶的複製

typedef struct complexnodecomplexnode;
complexnode* createcomplexnode(linknodetype value)
size_t diff(complexnode* src , complexnode* dest)
++offset;
src=src->next;
} if(src==null)
return offset;
}complexnode* step(complexnode* head ,size_t offset)
if(i>=offset)
++i;
cur = cur->next;
} return null;
}complexnode* copycomplexlist(complexnode* head)
//先按照簡單鍊錶來進行複製
complexnode* new_head = null;
complexnode* new_tail = null;
complexnode* cur = head;
for(;cur != null;cur = cur->next)else 
} //遍歷舊鍊錶，找到每個鍊錶節點random指標相對與鍊錶頭部的偏移量
//遍歷新鍊錶，根據偏移量，設定新鍊錶的random
complexnode* new_cur = new_head;
for(cur = head;cur!=null;cur=cur->next,new_cur=new_cur->next)
//通過diff函式來計算兩個節點的偏移量
size_t offset = diff(head ,cur->random);
//通過step函式，相當與求出從new_head出發，走了offset步，到達的位置是誰
new_cur->random = step(new_head,offset);
} return new_head;
}

兩個有序鍊錶求交集

linknode* linklistunionset(linknode* head1 , linknode* head2) 

linknode* cur1 = head1;
linknode* cur2 = head2;
linknode* new_head = null;
linknode* new_tail = null;
while(cur1 != null && cur2 != null) else if(cur1->data > cur2->data) else else
cur1 = cur1->next;
cur2 = cur2->next;
} }return new_head;
}

int linklisthascrosswithcycle(linknode* head1 , linknode* head2) 

//求連個鍊錶的入口點
linknode* entry1 = linklistcycleentry(head1);
linknode* entry2 = linklistcycleentry(head2);
//如果兩個鍊錶都不帶環，則按照前面的方法判定
if(entry1 == null && entry2 == null) 
//如果乙個帶環乙個不帶環，則不相交
if((entry1 == null && entry2 != null)||(entry1 != null && entry2 == null)) 
//如果兩個鍊錶都帶環
//a）如果連個入口相同，說明相交，並且是環外相交
if(entry1 == entry2) 
//b）如果從乙個入口點出發，繞環一周，能到達第二個入口點說明也是相交，並且是環上相交
linknode* cur = entry1->next;
while(cur != entry1) 
cur = cur->next;
} return 0;
}

求可能帶環的兩個鍊錶相交的交點

linknode* linklisthascrosswithcyclepos(linknode* head1 , linknode* head2) 

//交點在環外
linknode* entry1 = linklistcycleentry(head1);
linknode* entry2 = linklistcycleentry(head2);
if(entry1 == entry2) 
for(cur2 = head2 ;cur2 != entry2;cur2 = cur2->next)
if(len1 > len2) 
while(cur1 != null && cur2 != null) 
cur1 = cur1->next;
cur2 = cur2->next;
}}else 
while(cur1 != null && cur2 != null)
cur1 = cur1->next;
cur2 = cur2->next;
}} }
//環上相交
if(entry1 != entry2)
}

測試**：

void testhascrosswithcyle()

void testunionset()
void printcomplexlist(complexnode* head ,const char* msg)
printf("\n");
for(cur=head;cur!=null;cur=cur->next)
printf("[%c] ",cur->random->data);
} printf("\n");
}void testcomplexlistcopy()

鏈表面試題2

2 反轉鍊錶 定義乙個函式，輸入乙個鍊錶的頭節點，反轉該鍊錶並輸出反轉後鍊錶的頭節點。示例 輸入 1 2 3 4 5 null 輸出 5 4 3 2 1 null class solution cur.next prev prev cur cur curnext return newhead 反轉鍊...

鏈表面試題

不改變鍊錶結構，從尾到頭列印單鏈表 遞迴實現 void printlistrevers recursively plist phead printf d phead data 當鍊表非常長的時候，遞迴實現的會導致函式呼叫層級很深，可能導致呼叫棧溢位。用棧不會出現此類情況，顯然用棧實現 的魯棒性會好一...

鏈表面試題

從尾到頭列印單鏈表 棧 遞迴 從尾到頭列印鍊錶 includevoid printtailtohead listnode head while s.empty 空間複雜度o n void printtailtoheadr listnode head cout data 刪除乙個無頭單鏈表的非尾結點 ...