链表 典型习题

160 相交链表：遍历，统计是否出现过

/**
 * 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) {
        // 先把 A 链表的所有结点都访问一遍
        unordered_set visited;
        ListNode* temp = headA;
        while (temp != nullptr) {
            visited.insert(temp);
            temp = temp->next;
        }

        temp = headB;
        while (temp != nullptr) {
            if (visited.count(temp)) {
                return temp;
            }
            visited.insert(temp);
            temp = temp->next;
        }
        return nullptr;
        
    }
};

206 翻转链表：属于链表的基本操作之一

/**
 * 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* prev = nullptr;
        ListNode* curr = head;
        while (curr) {
            ListNode* next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;
        }
        return prev;

    }
};

234 回文链表

/**
 * 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:
    bool isPalindrome(ListNode* head) {
        vector vals;
        while (head != nullptr) {
            vals.push_back(head->val);
            head = head->next;
        }
        for (int i = 0, j = (int)vals.size() - 1; i < j; ++i, --j) {
            if(vals[i] != vals[j]) {
                return false;
            }
        }
        return true;
    }
};

141 环形链表

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool hasCycle(ListNode *head) {

        if (head == NULL || head->next == NULL) return false;

        ListNode* fast = head->next;
        ListNode* slow = head;

        while (slow != fast) {
            if (fast == NULL || fast->next == NULL) return false;
            slow = slow->next;
            fast = fast->next->next;
        }


        return true;
    }
};

142: 环形链表找起点：相遇之后复位再出发

a + n(b + c) = 2(a + b)

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode* fast = head;
        ListNode* slow = head;
        while (true) {
            if (fast == nullptr || fast->next == nullptr) return nullptr;
            fast = fast->next->next;
            slow = slow->next;
            if (fast == slow) break;
        }

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

21: 合并两个有序链表

/**
 * 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* mergeTwoLists(ListNode* l1, ListNode* l2) {
        // 建立头节点
        ListNode* preHead = new ListNode(-1);
        // 建立 prev
        ListNode* prev = preHead;
        // 添加更小的元素进来
        while (l1 != nullptr && l2 != nullptr) {
            if (l1->val < l2->val) {
                prev->next = l1;
                l1 = l1->next;
            } else {
                prev->next = l2;
                l2 = l2->next;
            }
            prev = prev->next;
        }

        prev->next = l1 == nullptr ? l2 : l1;

        return preHead->next;
    }
};

2.两数相加

/**
 * 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) {
        // 巧用哑节点
        ListNode* dummy = new ListNode(0, head);
        ListNode* second = dummy;
        ListNode* fast = head;

        for (int i = 0; i < n; ++i) {
            fast = fast->next;
        }

        while (fast!=nullptr) {
            second = second->next;
            fast = fast->next;
        }
        second->next = second->next->next;
        ListNode* ans = dummy->next;
        delete dummy;
        return ans;

    }
};

25 K 个一组翻转链表

/**
 * 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:
    // head 和 tail 是实际上需要被翻转的链表的头节点和尾节点
    pair myRev(ListNode* head, ListNode* tail) {
        ListNode* prev = tail->next;
        ListNode* p = head;

        while (prev != tail) {
            ListNode* nex = p->next;
            p->next = prev;
            prev = p;
            p = nex;
        }
        return {tail, head};

    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* hair = new ListNode(0, head);
        ListNode* pre = hair;

        while (head) {
            // 移动 tail k 步，到待翻转链表的尾部
            ListNode* tail = pre;
            for (int i = 0; i < k; i++) {
                tail = tail->next;
                if (!tail) {
                    // 不足以翻转，则结束
                    return hair->next;
                }
            }

            // 储存下一个节点，pre  已经有了
            ListNode* nex = tail->next;
            // 翻转
            pair result = myRev(head, tail);
            // 取头和尾
            head = result.first;
            tail = result.second;

            // 链接
            pre->next = head;
            tail->next = nex;
            // 移动
            pre = tail;
            head = tail->next;
        }

        return hair->next;
    }
};

146 LRU 缓存

# 定义一个存储数据的类
class DLinkedNode:
    def __init__(self, key = 0, value = 0):
        # 为什么要 key 和 value
        # 只有一个可以不？
        self.key = key
        self.value = value
        # 之前前驱和后继的两个指针
        self.prev = None
        self.next = None

class LRUCache:
    def __init__(self, capacity: int):
        # 初始化容量
        self.capacity = capacity
        self.size = 0

        # 初始化头尾
        self.head = DLinkedNode()
        self.tail = DLinkedNode()
        # 初始化两者的关系
        self.head.next = self.tail
        self.tail.prev = self.head

        # 一个字典储存现在的数据
        self.cache = dict()


    def get(self, key: int) -> int:

        if key not in self.cache:
            return -1
        
        # 取值
        node = self.cache[key]
        # 把读取过的节点移动到头部
        self.moveToHead(node)
        return node.value


    def put(self, key: int, value: int) -> None:

        if key not in self.cache:
            # 生成新的节点
            node = DLinkedNode(key, value)
            # 赋值
            self.cache[key] = node
            # 把新加入的数据直接放到头部有
            self.addToHead(node)
            # 修改大小
            self.size += 1

            if self.size > self.capacity:
                removed = self.removeTail();
                self.cache.pop(removed.key)
                self.size -= 1
        else:
            # 取值
            node = self.cache[key]
            # 赋值
            node.value = value
            # 把读取过的节点移动到头部
            self.moveToHead(node)
    
    def addToHead(self, node):
        node.prev = self.head
        node.next = self.head.next
        self.head.next.prev = node
        self.head.next = node

    def removeNode(self, node):
        node.prev.next = node.next
        node.next.prev = node.prev

    def moveToHead(self, node):
        # 先把节点摘掉
        self.removeNode(node)
        self.addToHead(node)

    def removeTail(self) -> DLinkedNode:
        node = self.tail.prev
        self.removeNode(node)
        return node





# Your LRUCache object will be instantiated and called as such:
# obj = LRUCache(capacity)
# param_1 = obj.get(key)
# obj.put(key,value)