leetcode 链表类型题总结

链表测试框架示例：

// leetcodeList.cpp : 定义控制台应用程序的入口点。vs2013 测试通过
//

#include "stdafx.h"
#include <Windows.h>
#include <iostream>

using namespace std;

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(nullptr){};
};

void showList(ListNode *Head) {
    ListNode *p = Head;
    while (p != nullptr) {
        cout << p->val << '\t';
        p = p->next;
    }
    cout << endl;
}

ListNode* createList(int n) {
    ListNode dummy(-1);
    ListNode *prev = &dummy;
    for (int i = 0; i < n; ++i) {
        int value;
        cout << "input number:";
        cin >> value;
        prev->next = new ListNode(value);
        prev = prev->next;
    }
    return dummy.next;
}

ListNode* addTwoNumbers(ListNode *l1, ListNode *l2) {
    ListNode *pa = l1;
    ListNode *pb = l2;
    ListNode dummy(-1);  // 头节点,目的是只有一个元素的时候不需要特殊考虑
    ListNode *prev = &dummy;
    int carry = 0; // 表示进位
    while (pa != nullptr || pb != nullptr) {
        const int ai = (pa == nullptr ? 0 : pa->val);
        const int bi = (pb == nullptr ? 0 : pb->val);
        const int value = (ai + bi + carry) % 10;
        carry = (ai + bi + carry) / 10;
        prev->next = new ListNode(value);

        pa = pa->next;
        pb = pb->next;
        prev = prev->next;
    }
    if (carry > 0)
        prev->next = new ListNode(carry);
    return dummy.next;
}


int _tmain(int argc, _TCHAR* argv[])
{
    ListNode *l1 = createList(3);
    showList(l1);
    ListNode *l2 = createList(3);
    showList(l2);
    ListNode *result = addTwoNumbers(l1, l2);
    showList(result);
    system("pause");
    return 0;
}

 

1, addTwoNumbers （思路同 数组题总结中的 plusOne 相似）

ListNode* addTwoNumbers(ListNode *l1, ListNode *l2) { 
    ListNode *pa = l1;
    ListNode *pb = l2;
    ListNode dummy(-1);  // 头节点,目的是只有一个元素的时候不需要特殊考虑
    ListNode *prev = &dummy;
    int carry = 0; // 表示进位
    while (pa != nullptr || pb != nullptr) {
        const int ai = (pa == nullptr ? 0 : pa->val);
        const int bi = (pb == nullptr ? 0 : pb->val);
        const int value = (ai + bi + carry) % 10;
        carry = (ai + bi + carry) / 10;
        prev->next = new ListNode(value);

        pa = pa->next;
        pb = pb->next;
        prev = prev->next;
    }
    if (carry > 0)
        prev->next = new ListNode(carry);
    return dummy.next;
}

 2，reverseBetween

ListNode* reverseBetween(ListNode *head, int m, int n) {
    ListNode dummy(-1);
    dummy.next = head;

    ListNode *prev = &dummy;
    for(int i = 0; i < m-1; ++i) {
        prev = prev->next;    // prev 指向第 m 个位置之前的元素
    }
    ListNode * head2 = prev;

    prev = head2->next;
    ListNode *cur = prev->next;     // cur 指向要执行头插操作的节点
    for (int i = m; i < n; ++i) {      // 头插的元素个数为 n-m 个
        prev->next = cur->next;
        cur->next = head->next;    // 头插法,从要反转位置的下一个元素进行头插
        head->next = cur;  

        cur = prev->next;
    }
    return dummy.next;
}

 3，partition List

ListNode* partition1(ListNode* head, int x) {  // *******这个方法特别好用*******
    ListNode left_dummy(-1);  // 定义小于3 的链表头节点
    ListNode right_dummy(-1);  // 定义大于等于3 的链表头节点

    auto left_cur = &left_dummy;
    auto right_cur = &right_dummy;

    for (ListNode *cur = head; cur != nullptr; cur = cur->next) {
        if (cur->val < x) {
            left_cur->next = cur;
            left_cur = cur;
        }
        else
        {
            right_cur->next = cur;
            right_cur = cur;
        }
    }
    left_cur->next = right_dummy.next;
    right_cur->next = nullptr;

    return left_dummy.next;
}


ListNode* partition2(ListNode *head, int x) {
    if (head == nullptr) return head;
    ListNode dummy(-1);
    dummy.next = head;

    ListNode *prev = &dummy;
    for (ListNode *curr = head; curr->next != nullptr; curr = curr->next) {  //对每一个元素进行判断，是否需要类头插
        while (curr->next != nullptr)
        {
            if (curr->val < x)
            {
                prev = curr;  // prev 始终指向最后一个小于 3 的节点，在此之后插入值小于 3 的节点
                curr = curr->next;
            }
            else
            {
                break;;
            }
        } 
        ListNode *prev_last = curr;  // 找到下一个要头插的节点
        ListNode *last = curr->next;
        while (last->next != nullptr)
        {
            if (last->val >= 3) {
                prev_last = last;
                last = last->next;
            }
            else
            {
                break;
            }
        }
        prev_last->next = last->next;  // 在最后一个小于 3 的元素之后插入
        last->next = prev->next;
        prev->next = last;   // 每次交换之后 curr 是指向最后一个大于等于 3 的节点，在此之前插入后面小于 3 的节点
    }
    return dummy.next;
}

 4, deleteDuplicates(I)

// ***********方法1*********************
ListNode* deleteDuplicates1(ListNode *head) {
    if (head == nullptr) return head;
    ListNode dummy(head->val + 1);  // 保证不和 head->val 相同即可
    dummy.next = head;

    ListNode *prev = &dummy;
    while (head != nullptr)       // head 指向后一个值，只要 head 有值，就要进行比较
    {
        if (prev->val == head->val) {
            prev->next = head->next;
            delete head;        // 注意内存泄漏，删除的节点要释放内存,temp 指向要删除的节点
            head = prev->next;  // 此时 head 为一个野指针，可以重新赋值
            
        }
        else {
            prev = head;
            head = head->next;
        }
    } 
    return dummy.next;
}

// ***********方法2*********************
void recur(ListNode *prev, ListNode *cur) {
    if (cur == nullptr) return;

    if (prev->val == cur->val) {
        prev->next = cur->next;
        delete cur;
        recur(prev, prev->next);
    }
    else {
        recur(prev->next, cur->next);
    }
}
ListNode* deleteDuplicates2(ListNode *head) {   // 递归版本
    if (!head) return head;
    ListNode dummy(head->val + 1);  // 理由同上
    dummy.next = head;

    recur(&dummy, head);  // 前一个和后一个
    return dummy.next;
}

// ***********方法3*********************
ListNode* deleteDuplicates3(ListNode *head){   // 迭代版本,基本思想和 deleteDuplicates1 一样
    if (head == nullptr) return nullptr;
    
    for (ListNode *prev = head, *curr = head->next; curr != nullptr; curr = prev->next) {
        if (prev->val == curr->val) {
            prev->next = curr->next;
            delete curr;   // 删除了 curr,系统并不会把 curr 设为空指针，而是还指向原来已经释放空间的内存地址
        }
        else {
            prev = curr;
        }
    }
    return head;
}

     deleteDuplicates(II)

ListNode* deleteDuplicatesII1(ListNode *head) {  // 迭代版本
    if (head == nullptr) return head;
    ListNode dummy(-1);  
//    dummy.next = head;

    ListNode *prev = &dummy;
    ListNode *curr = head;
    while (curr != nullptr)
    {
        bool duplicated = false;
        while (curr->next != nullptr && curr->val == curr->next->val) {
            duplicated = true;
            ListNode *temp = curr;
            curr = curr->next;
            delete temp;
        }
        if (duplicated) {  // 已经有重复元素，需要删除重复元素的最后一个元素
            ListNode *temp = curr;
            curr = curr->next;
            delete temp;
            continue;     // 继续检查下一个元素，重新回到起点判断。
        }
        prev->next = curr;
        prev = prev->next;
        curr = curr->next;
    }
    prev->next = curr;
    return dummy.next;
}

ListNode* deleteDuplicatesII2(ListNode *head) {  // 递归版本
    if (!head || head->next) return head;

    ListNode *p = head->next;
    if (head->val == p->val) {
        while (p && head->val == p->val) {
            ListNode *tmp = p;
            p = p->next;
            delete tmp;
        }
        delete head;
        return deleteDuplicatesII2(p);
    }
    else {
        head->next = deleteDuplicatesII2(head->next);
        return head;
    }
}

 5, rotate List

ListNode* rotateRight(ListNode *head, int k) {
    if (head == nullptr || k == 0) return head;

    int len = 1;
    ListNode *p = head;
    while (p->next != nullptr) {  // 求链表长度
        ++len;
        p = p->next;
    }
    k = len - k % len;
    p->next = head;
    for (int i = 1; i < k; ++i) {  // head 向后移动 k-1 步 | p 向右移动 k 步
        head = head->next;
    }
    p = head;         // 重新使用 p 指针
    head = head->next;
    p->next = nullptr;
    return head;
}

 6, remove Nth Node From End of List

ListNode* removeNthFromEnd(ListNode *head, int n) {
    if (head == nullptr) return head;
    ListNode dummy(-1);
    dummy.next = head;
    ListNode *first = &dummy;
    ListNode *last = &dummy;
    for (int i = 0; i < n; ++i) {  // 利用前后指针
        last = last->next;
    }
    while (last->next != nullptr) {
        first = first->next;
        last = last->next;
    }
    ListNode *temp = first->next;
    first->next = first->next->next;
    delete temp;
    return dummy.next;
}

 7, Swap Nodes in Pairs

ListNode* swapPairs1(ListNode *head) {
    if (!head || !head->next) return head;
    ListNode *p = head;
    ListNode dummy(-1);
    dummy.next = head->next;  // 直接指向新的头节点
    ListNode *prev = &dummy;

    while (p && p->next) {
        prev->next = p->next;
        ListNode *temp = p->next->next;
        p->next->next = p;
        p->next = temp;
        prev = p;
        p = temp;
    }
    return dummy.next;
}

ListNode* swapPairs2(ListNode *head) {
    ListNode* p = head;
    while (p && p->next) {
        swap(p->val, p->next->val);
        p = p->next->next;
    }
    return head;
}

 8, Reverse Nodes in k-Group

ListNode* reverseKGroup(ListNode *head, int k) {
    if (!head ||!head->next || k <= 1) return head;
    ListNode dummy(-1);
    dummy.next = head;
    ListNode *prev = &dummy;
    bool enough = true;

    while (true) {
        ListNode *p = head;
        ListNode *begin = head;
        for (int i = 0; i < k; ++i) {
            if (p) p = p->next;
            else {
                enough = false;
                prev->next = begin;   // 把最后不够反转的节点连接起来
                return dummy.next;
            }
        }
        ListNode *prev_next = begin;
        if (enough) {
            for (int i = 0; i < k; ++i) { // 头插法
                ListNode* temp = begin->next;
                if (i == 0) begin->next = nullptr;   // 如果是最后一个节点，就把该节点的 next 置为 nullptr ,为了 showList 函数能展示
                else       begin->next = prev->next;
                prev->next = begin;
                begin = temp;
            }
        }
//        showList(dummy.next);  // using test
        prev = prev_next;
        head = begin;
    }// while
}

 9,  Linked List Cycle

bool hasCycle1(ListNode *head) {  // STL::unordered_map
    unordered_map<ListNode*, bool> visted;  // 检测节点地址是否重复访问
    for (head; head != nullptr; head = head->next) {
        if (visted.find(head) != visted.end()) {
            visted.find(head)->second = true;   // 有没有都可以
            return false;
        }
        else {
            visted.find(head)->second = false;
        }    
    }
    return true;
}

bool hasCycle2(ListNode *head) {
    ListNode *slow = head;
    ListNode *fast = head;
    while (fast && fast->next) {
        slow = slow->next;       // slow 每次走一步
        fast = fast->next->next; // fast 每次走两步，看作 fast 追 slow 的话，每次距离缩短 1 ，不会跳过slow的
        if (slow == fast)
            return true;
    }
    return false;
}

      Linked List Cycle(II)

ListNode* detectCycle1(ListNode *head) {  //  STL::unordered_map
    unordered_map<ListNode*, bool> visted;
    for (head; head != nullptr; head = head->next) {
        if (visted.find(head) != visted.end()) {
            return head;
        }
        else {
            visted.find(head)->second = false;
        }
    }
    return nullptr;
}

ListNode* detectCycle2(ListNode *head) {
    ListNode* slow = head;
    ListNode* fast = head;

    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
        if (slow == fast) {
            ListNode *p = head;
            while (p != slow) {   //  思路来源：https://blog.csdn.net/xy010902100449/article/details/48995255
                slow = slow->next;
                p = p->next;
            }
            return p;
        }
    }
    return nullptr;
}

 10, Reorder List

ListNode* reverse(ListNode *head) {
    if (!head || !head->next) return head;
    ListNode dummy(-1);
    dummy.next = head;

    ListNode *prev = &dummy;
    ListNode *curr = head;
    while (curr) {
        ListNode* temp = curr->next; // 保存下一个节点
        if (curr == head) curr->next = nullptr;    // 头插法
        else              curr->next = prev->next;
        prev->next = curr;
        curr = temp;
    }
    return dummy.next;
}

ListNode* mergeList(ListNode *list1, ListNode *list2) {  // 把 list2 合并到 list1 中
    if (!list1) return list2;
    if (!list2) return list1;

    ListNode *curr1 = list1;
    ListNode *curr2 = list2;
    ListNode *temp1 = nullptr;
    ListNode *temp2 = nullptr;
    while (curr1->next && curr2->next) {
        temp1 = curr1->next;
        temp2 = curr2->next;
        curr2->next = curr1->next;
        curr1->next = curr2;
        curr1 = temp1;
        curr2 = temp2;
    }
    if (!curr1->next)  // 如果第一个链表节点少，需要把长的链表后面的节点链接上，如果第二个少，不需要做处理
        curr1->next = temp2;
    return list1;
}

ListNode* reorderList(ListNode *head) {  // 先反转后半部分链表，然后再交互合并两个链表
    if (!head || !head->next) return head;

    ListNode *slow = head;
    ListNode *fast = head;
    ListNode *prev = nullptr;

    while (fast && fast->next) {
        prev = slow;           // prev 后面断开,slow 指向后半链表的第一个节点
        slow = slow->next;
        fast = fast->next->next;
    }
    prev->next = nullptr;
    ListNode *head2 = reverse(slow);

    // merge two lists head && head2
    head = mergeList(head, head2);  //  把 head2 合并到 head 中
    return head;
}

 11, LRU Cache

class LRUCache {
private:
    struct CacheNode {
        int key;
        int value;
        CacheNode(int k, int v) : key(k), value(v) {};
    };
public:
    LRUCache(int capacity) {
        this->capacity = capacity;
    }
    int get(int key) {
        if (cacheMap.find(key) == cacheMap.end()) return -1;

        cacheList.splice(cacheList.begin(), cacheList, cacheMap[key]);  // 移动最新访问的节点到最前面
        cacheMap[key] = cacheList.begin();  // 更新 map 中节点的地址
        return cacheMap[key]->value;
    }
    void set(int key, int value) {
        if (cacheMap.find(key) == cacheMap.end()) { // 没有该元素
            if (cacheList.size() == capacity) {  // 空间已满
                cacheMap.erase(cacheList.back().key);
                cacheList.pop_back();
            }
            // 空间不满，插入新节点到链表头部
            cacheList.push_front(CacheNode(key, value));
            cacheMap[key] = cacheList.begin();
        }
        else {  // 存在该元素，更新节点值，并把该节点放到链表头部，更新 cacheMap 中的值
            cacheMap[key]->key = value;
            cacheList.splice(cacheList.begin(), cacheList, cacheMap[key]);
            cacheMap[key] = cacheList.begin();
        }
    }
private:
    list<CacheNode> cacheList;
    unordered_map<int, list<CacheNode>::iterator> cacheMap;
    int capacity;
};

 12, deep Copy

RandomListNode* copyRandomList1(RandomListNode *head) { // STL::unordered_map
    if (head == nullptr) return nullptr;

    RandomListNode dummy(-1);
    RandomListNode *scan = &dummy;
    RandomListNode *curr = head;
    unordered_map<RandomListNode*, RandomListNode*> mapping;

    while (curr != nullptr) {
        scan->next = new RandomListNode(curr->label);
        mapping.insert(curr, scan->next);  // 存储 原链表节点和复制链表节点的对应关系
        curr = curr->next;
        scan = scan->next;
    }
    curr = head;
    scan = &dummy;
    while (curr != nullptr) {  // 赋值各个节点的 random 域
        scan->next->random = mapping[curr->random];
        curr = curr->next;
        scan = scan->next;
    }
    return dummy.next;
}

RandomListNode* copyRandomList2(RandomListNode *head) {
    if (head == nullptr) return head;

    RandomListNode *scan = head;
    while (!scan) {   // step1: 第一遍扫描，对每个节点赋值
        RandomListNode *newNode = new RandomListNode(scan->label);
        newNode->next = scan->next;
        scan->next = newNode;
        scan = newNode->next;
    }
    scan = head;  // step2: 第2遍扫描，对每个复制节点的 random 域进行赋值
    while (!scan) {
        if (!scan->random) {
            scan->next->random = scan->random->next;  // 把前面复制节点的 random 指向后面复制节点的 random
        }
    }
    RandomListNode *newHead = head->next;
    scan = head;   // step3:第3遍扫描，拆分奇偶节点，分成两个链表，奇数组成的链表为原链表，偶数组成的链表为复制链表
    while (!scan) {
        RandomListNode *evenNode = scan->next;
        scan->next = evenNode->next;
        if (scan->next != nullptr) {
            evenNode->next = scan->next->next;
        }
        scan = scan->next;
    }
    return newHead;
}

 

 

 

以上题目来源于：https://github.com/soulmachine/leetcode（leetcode-cpp.pdf）