• 今日学习的文章链接，或者视频链接

第二章 链表part02

• 自己看到题目的第一想法

• 看完代码随想录之后的想法

24:

注意链表的操作：

class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        auto dummyhead = new ListNode(0,head);
        auto prev = dummyhead;
        auto cur = head;
        while (cur != nullptr&&cur->next != nullptr){
            auto tmp = cur->next;
            auto tmp_next = tmp->next;
            prev -> next = tmp;
            tmp->next = cur;
            cur->next = tmp_next;
            prev = cur;
            cur = tmp_next;
        }
        return dummyhead->next;
    }
};

19：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* dummyhead = new ListNode(-1,head);
        ListNode* fast = dummyhead;
        ListNode* slow = dummyhead;
        for(int i = 0; i < n+1; i++){
            fast = fast->next;
        }
        while(fast != nullptr){
            fast = fast -> next;
            slow = slow -> next;
        }
        slow -> next = slow -> next -> next;
        return dummyhead->next;
    }
};

不用dummyhead版本，各种处理边界条件：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        //ListNode* dummyhead = new ListNode(-1,head);
        ListNode* fast = head;
        ListNode* slow = head;
        int flag = 1;
        for(int i = 0; i < n+1; i++){
            if (fast != nullptr)
            {
                fast = fast->next;
            }else{
                flag = 0;
                break;
            }
        }
        while(fast != nullptr){
            fast = fast -> next;
            slow = slow -> next;
        }
        if (flag==1||slow!=head&&slow -> next != nullptr){
            slow -> next = slow -> next -> next;
        }else{
            return head->next;
        }       
        return head;
    }
};

02.07

奇妙解法：

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        auto curA = headA;
        auto curB = headB;
        while (curA != curB){
            if (curA == nullptr){
                curA = headB;
            }else{
                curA = curA->next;
            }
            curB = curB == nullptr?headA:curB->next;
        }
        return curA;
    }
};

142：

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        auto fast = head;
        auto slow = head;
        //fast 走两步，所以判断fast和fast->next
        while (fast != nullptr && fast->next != nullptr) {
            fast = fast->next->next;
            slow = slow->next;
            if (fast == slow) {
                break;
            }
        }
        if (fast == nullptr || fast->next == nullptr) {
            return nullptr; // No cycle found
        }
        fast = head;
        while(fast!=slow){
            fast = fast->next;
            slow = slow->next;
        }
        return fast;
    }
};

 

• 自己实现过程中遇到哪些困难

• 今日收获，记录一下自己的学习时长