1、比較順序表和煉表的優缺點,它們分別在什麼場景下使用?
1)順序表支援隨機訪問,單鏈表不支援隨機訪問。
2)順序表插入/刪除資料效率很低,時間複雜度為o(n)(除尾插和尾刪),單鏈表插入/刪除效率更高,時間複雜度為o(1)。
3)順序表的cpu高速緩衝效率更高,單鏈表cpu高速緩衝效率低。
2、列印單向鍊錶
void display(plist plist)
printf("over\n");
}
3、單鏈表的排序
void bubblesort(plist *pplist)
cur =
*pplist;
while (cur != tail)
cur = cur->next;
}tail = cur;
cur =
*pplist;
}}
4、逆序列印單項鍊表
void reverseprint(plist plist)
if (cur->next !=
null)
printf("%3d", cur->
data);
}
5、刪除無頭單鏈表的非尾結點
void erasenottail(pnode pos)
6、在無頭單鏈表的非頭結點前插入乙個元素
void insertfrontnode(pnode pos, datatype x)
7、約瑟夫環問題
void josephcycle(plist *pplist, int k)
del = cur->next;
printf("%d", cur->
data);
cur->
data
= cur->next->
data;
cur->next = cur->next->next;
free(del);
del =
null;
}printf("%d\n", cur->
data);
}
8、逆序單向鍊錶
void reverselist(plist *pplist)
cur = *pplist;
next = cur->next;
cur->next = null;
while (next != null)
*pplist = cur;
}
9、合併兩個有序列表
plist merge(const plist *p1, const plist *p2)
if (list1 ==
null)
if (list2 ==
null)
if (list1->
data
< list2->
data)//新鍊錶的首部
else
tail = newlist;
while (list1 && list2)
else
}if (list1 =
null)
else
return newlist;
}
10、查詢單鏈表的中間節點,要求只能遍歷一次鍊錶
//定義兩個指標分別為fast和slow,fast每次跳兩下,slow每次跳一下,
//當fast指向鏈尾時,show所指的位置就是中間結點
pnode findmidnode(plist plist)
return slow;
}
11、查詢單鏈表的倒數第k個節點,要求只能遍歷一次鍊錶
//定義兩個指標分別為fast和slow,讓fast先走k-1步後slow再走,當fast指向尾部時,slow所指向的就是倒數第k個元素
pnode findknode(plist plist, int k)
fast = fast->next;
}if (k <=
0)//執行完while後才能輸出
return
null;
}
12、判斷鍊錶帶環情況
//定義兩個指標分別為fast和slow,每次fast比slow多走一步,如果帶環,那麼它們一定相交
pnode checkcircle(plist plist)
}return
null;
}
13、求環的長度
int getcirclelength(pnode meet)
return i;
}
14、求環的入口點
pnode getcycleentrynode(plist plist, pnode meet)
while(pnode != meet)
return meet;
}
15、判斷兩條單項鍊表是否相交
int checkcross(plist list1, plist list2)
while (p2 && p2->next)
if ((p1 == p2) && (p1 !=
null))//相交
else
return-1;
}
16、轉化成不帶環的鍊錶求解
pnode getcrossnocycle(plist list1, plist list2)
cur->next = list2;
meet = checkcircle(list1);
crossnode = getcycleenternode(list1, meet);
return crossnode;
}
17、求交點
pnode getcrossnode(plist list1, plist list2)
else
if (p1 && p2)
return crossnode;
}
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