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链表

203.移除链表元素​

力扣题目链接(opens new window)

题意：删除链表中等于给定值 val 的所有节点。

示例 1： 输入：head = [1,2,6,3,4,5,6], val = 6 输出：[1,2,3,4,5] 示例 2： 输入：head = [], val = 1 输出：[] 示例 3： 输入：head = [7,7,7,7], val = 7 输出：[]

笔记：

  • 创建一个虚拟头节点 (dummy head)，更好删除。
  • Rust 感觉是最复杂的，学习了 as_mut() 和 take() 这两个 API，每次要删除之前，直接使用 take() 转移了 Next 节点的所有权，如果不用删除再放回去。

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode *dummy_head = new ListNode();
        dummy_head->next = head;
        ListNode *current_node = dummy_head;
        while (current_node->next != nullptr) {
            if (current_node->next->val == val) {
                ListNode *to_be_deleted = current_node->next;
                current_node->next = current_node->next->next;
                delete to_be_deleted;
            } else {
                current_node = current_node->next;
            }
        }
        head = dummy_head->next;
        delete dummy_head;
        return head;
    }
};

Rust

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn remove_elements(head: Option<Box<ListNode>>, val: i32) -> Option<Box<ListNode>> {
        let mut dummy_head = Box::new(ListNode::new(0));
        dummy_head.next = head;
        let mut current_node = dummy_head.as_mut();
        while let Some(next_node) = current_node.next.take() {
            if next_node.val == val {
                current_node.next = next_node.next;
            } else {
                current_node.next = Some(next_node);
                current_node = current_node.next.as_mut().unwrap();
            }
        }
        dummy_head.next
    }
}

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func removeElements(head *ListNode, val int) *ListNode {
    dummyHead := &ListNode{}
    dummyHead.Next = head;
    current := dummyHead
    for current.Next != nil {
        if current.Next.Val == val {
            current.Next = current.Next.Next
        } else {
            current = current.Next
        }
    }
    return dummyHead.Next
}

707.设计链表​

力扣题目链接(opens new window)

题意：

在链表类中实现这些功能：

  • get(index)：获取链表中第  index  个节点的值。如果索引无效，则返回-1。
  • addAtHead(val)：在链表的第一个元素之前添加一个值为  val  的节点。插入后，新节点将成为链表的第一个节点。
  • addAtTail(val)：将值为  val 的节点追加到链表的最后一个元素。
  • addAtIndex(index,val)：在链表中的第  index  个节点之前添加值为  val  的节点。如果  index  等于链表的长度，则该节点将附加到链表的末尾。如果 index 大于链表长度，则不会插入节点。如果 index 小于 0，则在头部插入节点。
  • deleteAtIndex(index)：如果索引  index 有效，则删除链表中的第  index 个节点。

笔记：

  • 需要注意的细节太多了，什么时候使用虚拟头节点，什么时候使用头节点
  • 明天不做新题，实现双向链表，Rust 先尝试 safe 实现

C++

  • 单链表实现
class MyLinkedList {
public:
    struct LinkedNode {
        int val_;
        LinkedNode *next_;
        LinkedNode(int val): val_(val), next_(nullptr) {}
    };

    MyLinkedList() {
        dummy_head_ = new LinkedNode(0);
        size_ = 0;
    }

    int get(int index) {
        if (index >= size_ || index < 0) {
            return -1;
        }
        LinkedNode *current = dummy_head_->next_;
        while (index > 0) {
            current = current->next_;
            index--;
        }
        return current->val_;
    }

    void addAtHead(int val) {
        LinkedNode *new_node = new LinkedNode(val);
        new_node->next_ = dummy_head_->next_;
        dummy_head_->next_ = new_node;
        size_++;
    }

    void addAtTail(int val) {
        LinkedNode *new_node = new LinkedNode(val);
        LinkedNode *current = dummy_head_;
        while (current->next_ != nullptr) {
            current = current->next_;
        }
        current->next_ = new_node;
        size_++;
    }

    void addAtIndex(int index, int val) {
        if (index > size_ || index < 0) {
            return;
        }
        LinkedNode *new_node = new LinkedNode(val);
        LinkedNode *current = dummy_head_;
        while (index > 0) {
            current = current->next_;
            index--;
        }
        new_node->next_ = current->next_;
        current->next_ = new_node;
        size_++;
    }

    void deleteAtIndex(int index) {
        if (index >= size_ || index < 0) {
            return;
        }
        LinkedNode *current = dummy_head_;
        while (index > 0) {
            current = current->next_;
            index--;
        }
        LinkedNode *to_be_deleted = current->next_;
        current->next_ = current->next_->next_;
        delete to_be_deleted;
        size_--;
    }
private:
    int size_;
    LinkedNode *dummy_head_;
};

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * MyLinkedList* obj = new MyLinkedList();
 * int param_1 = obj->get(index);
 * obj->addAtHead(val);
 * obj->addAtTail(val);
 * obj->addAtIndex(index,val);
 * obj->deleteAtIndex(index);
 */
  • 双向链表实现 ^c06900
class MyLinkedList {
    struct LinkedNode {
        int val;
        LinkedNode *prev;
        LinkedNode *next;

        LinkedNode(int val) :val(val), prev(nullptr), next(nullptr){}
    };

public:
    MyLinkedList() {
        dummy = new LinkedNode(0);
        dummy->next = dummy;
        dummy->prev = dummy;
        size = 0;
    }

    int get(int index) {
        if (index >= size || index < 0) {
            return -1;
        }
        LinkedNode *curr = dummy->next;
        while (index > 0) {
            curr = curr->next;
            index--;
        }
        return curr->val;
    }

    void addAtHead(int val) {
        LinkedNode *new_node = new LinkedNode(val);
        new_node->next = dummy->next;
        new_node->next->prev = new_node;
        dummy->next = new_node;
        new_node->prev = dummy;
        size++;
    }

    void addAtTail(int val) {
        LinkedNode *new_node = new LinkedNode(val);
        new_node->prev = dummy->prev;
        new_node->next = dummy;
        dummy->prev = new_node;
        new_node->prev->next = new_node;
        size++;
    }

    void addAtIndex(int index, int val) {
        if (index > size || index < 0) {
            return;
        }
        LinkedNode *curr = dummy;
        while (index > 0) {
            curr = curr->next;
            index--;
        }
        LinkedNode *new_node = new LinkedNode(val);
        new_node->next = curr->next;
        new_node->prev = curr;
        curr->next = new_node;
        new_node->next->prev = new_node;
        size++;
    }

    void deleteAtIndex(int index) {
        if (index >= size || index < 0) {
            return;
        }
        LinkedNode *curr = dummy;
        while (index > 0) {
            curr = curr->next;
            index--;
        }
        LinkedNode *to_be_deleted = curr->next;
        curr->next = to_be_deleted->next;
        curr->next->prev = curr;
        delete to_be_deleted;
        size--;
    }

private:
    LinkedNode *dummy;
    int size;
};

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * MyLinkedList* obj = new MyLinkedList();
 * int param_1 = obj->get(index);
 * obj->addAtHead(val);
 * obj->addAtTail(val);
 * obj->addAtIndex(index,val);
 * obj->deleteAtIndex(index);
 */

Rust

Go

type LinkedNote struct {
    val int
    next *LinkedNote
}

type MyLinkedList struct {
    dummyHead *LinkedNote
    size int
}

func Constructor() MyLinkedList {
    list := MyLinkedList{new(LinkedNote), 0}
    return list
}

func (this *MyLinkedList) Get(index int) int {
    if index < 0 || index >= this.size {
        return -1
    }
    curr := this.dummyHead.next
    for i := 0; i < index; i++ {
        curr = curr.next
    }
    return curr.val
}

func (this *MyLinkedList) AddAtHead(val int)  {
    newNode := &LinkedNote{val: val}
    newNode.next = this.dummyHead.next
    this.dummyHead.next = newNode
    this.size++
}

func (this *MyLinkedList) AddAtTail(val int)  {
    newNode := &LinkedNote{val: val}
    curr := this.dummyHead
    for curr.next != nil {
        curr = curr.next
    }
    curr.next = newNode
    this.size++
}

func (this *MyLinkedList) AddAtIndex(index int, val int)  {
    if index < 0 || index > this.size {
        return
    }
    curr := this.dummyHead
    for i := 0; i < index; i++ {
        curr = curr.next
    }
    newNode := &LinkedNote{val, curr.next}
    curr.next = newNode
    this.size++
}

func (this *MyLinkedList) DeleteAtIndex(index int)  {
    if index < 0 || index >= this.size {
        return
    }
    curr := this.dummyHead
    for i := 0; i < index; i++ {
        curr = curr.next
    }
    curr.next = curr.next.next
    this.size--
}

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * obj := Constructor();
 * param_1 := obj.Get(index);
 * obj.AddAtHead(val);
 * obj.AddAtTail(val);
 * obj.AddAtIndex(index,val);
 * obj.DeleteAtIndex(index);
 */

206.反转链表​

力扣题目链接

题意：反转一个单链表。

示例: 输入: 1->2->3->4->5->NULL 输出: 5->4->3->2->1->NULL

儿童节，恢复打卡。

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode *temp_node = nullptr;
        ListNode *current = head;
        ListNode *prev = nullptr;
        while (current != nullptr) {
            temp_node = current->next;
            current->next = prev;
            prev = current;
            current = temp_node;
        }
        return prev;
    }
};

24. 两两交换链表中的节点​

给定一个链表，两两交换其中相邻的节点，并返回交换后的链表。

你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

思路：一开始比较疑惑剩下一个节点怎么办，发现这道题 21 年做过。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode* dummy_head = new ListNode();
        dummy_head->next = head;
        ListNode* current = dummy_head;

        while (current->next != nullptr && current->next->next != nullptr) {
            ListNode *first_node = current->next;
            ListNode *second_node = current->next->next;
            first_node->next = second_node->next;
            second_node->next = first_node;
            current->next = second_node;
            current = first_node;
        }
        return dummy_head->next;
    }
};

19. 删除链表的倒数第 N 个节点​

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

进阶：你能尝试使用一趟扫描实现吗？

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode *dummy_head = new ListNode();
        dummy_head->next = head;

        ListNode *fast = dummy_head;
        ListNode *slow = dummy_head;

        // fast go n steps
        while (fast->next != nullptr && n > 0) {
            fast = fast->next;
            n--;
        }

        while (fast->next != nullptr) {
            fast = fast->next;
            slow = slow->next;
        }

        ListNode *to_be_deleted = slow->next;
        slow->next = to_be_deleted->next;
        delete to_be_deleted;
        return dummy_head->next;
    }
};

160.链表相交​

力扣题目链接

给你两个单链表的头节点  headA 和 headB ，请你找出并返回两个单链表相交的起始节点。如果两个链表没有交点，返回 null 。

图示两个链表在节点 c1 开始相交。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *ptr_a = headA;
        ListNode *ptr_b = headB;

        int len_a = 0;
        int len_b = 0;

        while (ptr_a != nullptr) {
            len_a++;
            ptr_a = ptr_a->next;
        }
        while (ptr_b != nullptr) {
            len_b++;
            ptr_b = ptr_b->next;
        }

        if (len_a == 0 || len_b == 0) {
            return nullptr;
        }

        ptr_a = headA;
        ptr_b = headB;
        if (len_a >= len_b) {
            int diff = len_a - len_b;
            while (diff > 0) {
                diff--;
                ptr_a = ptr_a->next;
            }
        } else {
            int diff = len_b - len_a;
            while (diff > 0) {
                diff--;
                ptr_b = ptr_b->next;
            }
        }

        while (ptr_a != ptr_b) {
            ptr_a = ptr_a->next;
            ptr_b = ptr_b->next;
            if (ptr_a == nullptr || ptr_b == nullptr) {
                return nullptr;
            }
        }
        return ptr_a;
    }
};

142.环形链表 II​

力扣题目链接(opens new window)

题意： 给定一个链表，返回链表开始入环的第一个节点。 如果链表无环，则返回 null。

为了表示给定链表中的环，使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 pos 是 -1，则在该链表中没有环。

说明：不允许修改给定的链表。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode *fast = head;
        ListNode *slow = head;
        while (fast != nullptr && fast->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) {
                ListNode *p_head = head;
                ListNode *p_cross = slow;
                while (p_head != p_cross) {
                    p_head = p_head->next;
                    p_cross = p_cross->next;
                }
                return p_head;
            }
        }
        return nullptr;
    }
};