707. 设计链表

leetcode链接:
https://leetcode.cn/problems/design-linked-list/

方案一

注:此方案无法正常执行
原因:底层开始设计出现问题,细节决定成败

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class MyLinkedList {

int val;

MyLinkedList next;

MyLinkedList prev;

boolean numFlag;

public MyLinkedList() {
this.next = null;
this.prev = null;
}

public int get(int index) {
MyLinkedList temp = this;

for (; index > 0; index--) {
temp = temp.next;
if (temp == null) {
break;
}
}

if (temp == null) {
return -1;
} else {
return temp.val;
}
}

public void addAtHead(int val) {
if (!numFlag) {
this.val = val;
this.numFlag = true;
} else {

MyLinkedList temp = new MyLinkedList();
temp.prev = this;
temp.val = this.val;
temp.next = this.next;
temp.numFlag = this.numFlag;

if (this.next != null) {
this.next.prev = temp;
}

this.next = temp;
this.val = val;
}
}

public void addAtTail(int val) {
if (!numFlag) {
this.val = val;
this.numFlag = true;
} else {
MyLinkedList temp = this;
while(temp != null) {
if (temp.next == null){

temp.next = new MyLinkedList();
temp.next.prev = temp;
temp.next.val = val;
temp.next.numFlag = true;

break;
} else {
temp = temp.next;
}
}
}
}

public void addAtIndex(int index, int val) {

MyLinkedList temp = this;

for (; index > 0; index--) {
if (temp.next == null) {
break;
} else {
temp = temp.next;
}
}

if (index == 0) {
MyLinkedList newList = new MyLinkedList();
newList.val = temp.val;
newList.numFlag = temp.numFlag;
newList.next = temp.next;
newList.prev = temp;

temp.val = val;
temp.next = newList;
} else if (index == 1) {
MyLinkedList newList = new MyLinkedList();
newList.val =val;
newList.numFlag = true;
newList.prev = temp;

temp.next = newList;
}


}

public void deleteAtIndex(int index) {
MyLinkedList temp = this;

for (; index > 0; index--) {
temp = temp.next;
if (temp == null) {
break;
}
}

if (temp != null) {
if (temp.next != null) {
temp.next.prev = temp.prev;
} else {
if (temp.prev != null) {
temp.prev.next = null;
}
}

if (temp.prev != null) {
temp.prev.next = temp.next;
} else {
if (temp.next != null) {
temp.val = temp.next.val;
if (temp.next.next != null) {
temp.next.next.prev = temp;
}
temp.next = temp.next.next;
} else{
temp.val = 0;
temp.numFlag = false;
temp.next = null;
temp.prev=null;
}

}
}
}
}

Wrong Answer

  • 28/65 cases passed (N/A)

Testcase

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["MyLinkedList","addAtHead","addAtTail","addAtTail","addAtTail","addAtTail","addAtTail","addAtTail","deleteAtIndex","addAtHead","addAtHead","get","addAtTail","addAtHead","get","addAtTail","addAtIndex","addAtTail","addAtHead","addAtHead","addAtHead","get","addAtIndex","addAtHead","get","addAtHead","deleteAtIndex","addAtHead","addAtTail","addAtTail","addAtIndex","addAtTail","addAtHead","get","addAtTail","deleteAtIndex","addAtIndex","deleteAtIndex","addAtHead","addAtTail","addAtHead","addAtHead","addAtTail","addAtTail","get","get","addAtHead","addAtTail","addAtTail","addAtTail","addAtIndex","get","addAtHead","addAtIndex","addAtHead","addAtTail","addAtTail","addAtIndex","deleteAtIndex","addAtIndex","addAtHead","addAtHead","deleteAtIndex","addAtTail","deleteAtIndex","addAtIndex","addAtTail","addAtHead","get","addAtIndex","addAtTail","addAtHead","addAtHead","addAtHead","addAtHead","addAtHead","addAtHead","deleteAtIndex","get","get","addAtHead","get","addAtTail","addAtTail","addAtIndex","addAtIndex","addAtHead","addAtTail","addAtTail","get","addAtIndex","addAtHead","deleteAtIndex","addAtTail","get","addAtHead","get","addAtHead","deleteAtIndex","get","addAtTail","addAtTail"]
' +
'[[],[38],[66],[61],[76],[26],[37],[8],[5],[4],[45],[4],[85],[37],[5],[93],[10,23],[21],[52],[15],[47],[12],[6,24],[64],[4],[31],[6],[40],[17],[15],[19,2],[11],[86],[17],[55],[15],[14,95],[22],[66],[95],[8],[47],[23],[39],[30],[27],[0],[99],[45],[4],[9,11],[6],[81],[18,32],[20],[13],[42],[37,91],[36],[10,37],[96],[57],[20],[89],[18],[41,5],[23],[75],[7],[25,51],[48],[46],[29],[85],[82],[6],[38],[14],[1],[12],[42],[42],[83],[13],[14,20],[17,34],[36],[58],[2],[38],[33,59],[37],[15],[64],[56],[0],[40],[92],[63],[35],[62],[32]]

Answer

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[null,null,null,null,null,null,null,null,null,null,null,61,null,null,61,null,null,null,null,null,null,85,null,null,37,null,null,null,null,null,null,null,null,23,null,null,null,null,null,null,null,null,null,null,-1,23,null,null,null,null,null,31,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,8,null,null,null,null,null,null,null,null,null,6,47,null,39,null,null,null,null,null,null,null,93,null,null,null,null,83,null,93,null,null,59,null,null]

Expected Answer

1
[null,null,null,null,null,null,null,null,null,null,null,61,null,null,61,null,null,null,null,null,null,85,null,null,37,null,null,null,null,null,null,null,null,23,null,null,null,null,null,null,null,null,null,null,-1,95,null,null,null,null,null,31,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,8,null,null,null,null,null,null,null,null,null,6,47,null,23,null,null,null,null,null,null,null,93,null,null,null,null,48,null,93,null,null,59,null,null]

方案二

重构方案:
1、采用虚拟头,方便在头中插入数值
2、定义一个节点类,方便对其进行操作,逻辑更容易梳理

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class ListNode {
int val;
ListNode next;

public ListNode(int val) {
this.val = val;
}
}

作者:LeetCode-Solution
链接:https://leetcode.cn/problems/design-linked-list/solution/she-ji-lian-biao-by-leetcode-solution-abix/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

3、类中加入一个size标签,判断是否是否超出索引,将一部分判断异常逻辑剥离出来。
4、使用单链表建立。

参考来自于:
https://programmercarl.com/0707.%E8%AE%BE%E8%AE%A1%E9%93%BE%E8%A1%A8.html#%E4%BB%A3%E7%A0%81
https://www.bilibili.com/video/BV1FU4y1X7WD/

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class MyLinkedList {

int size;

ListNode listNode;

// MyLinkedList() 初始化 MyLinkedList 对象。
public MyLinkedList() {
size = -1;
listNode = new ListNode(0);

}

// int get(int index) 获取链表中下标为 index 的节点的值。如果下标无效,则返回 -1 。
public int get(int index) {
if (index < 0 || index > size) {
return -1;
}
ListNode tempNode = listNode;

while (index > 0) {
tempNode = tempNode.next;
index--;
}

return tempNode.next.val;
}

//void addAtHead(int val) 将一个值为 val 的节点插入到链表中第一个元素之前。在插入完成后,新节点会成为链表的第一个节点。
public void addAtHead(int val) {
addAtIndex(0, val);
}

// void addAtTail(int val) 将一个值为 val 的节点追加到链表中作为链表的最后一个元素。
public void addAtTail(int val) {
addAtIndex(size + 1, val);
}

// void addAtIndex(int index, int val) 将一个值为 val 的节点插入到链表中下标为 index 的节点之前。如果 index 等于链表的长度,那么该节点会被追加到链表的末尾。如果 index 比长度更大,该节点将 不会插入 到链表中。
public void addAtIndex(int index, int val) {
if (index < 0 || index > (size + 1)) {
return;
}
ListNode tempNode = listNode;

while (index > 0) {
tempNode= tempNode.next;
index--;
}

ListNode addNode = new ListNode(val);

addNode.next = tempNode.next;
tempNode.next = addNode;

this.size++;
}

// void deleteAtIndex(int index) 如果下标有效,则删除链表中下标为 index 的节点。
public void deleteAtIndex(int index) {
if (index < 0 || index > size) {
return;
}
ListNode tempNode = listNode;

while (index > 0) {
tempNode = tempNode.next;
index--;
}
tempNode.next = tempNode.next.next;
this.size--;
}

}

class ListNode {
int val;

ListNode next;

public ListNode(int val) {
this.val = val;
}
}

Accepted

  • 65/65 cases passed (9 ms)
  • Your runtime beats 87.87 % of java submissions
  • Your memory usage beats 19.66 % of java submissions (43.2 MB)

707. 设计链表
http://yuanql.top/2023/06/04/02_leetcode/707. 设计链表/
作者
Qingli Yuan
发布于
2023年6月4日
许可协议