LeetCode刷题笔记 206. 反转链表

题目要求

题解

头插法

简单讲就是从第二个元素起依次插到链表的头部。

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(!head||!head->next) return head;
        ListNode dummy(-1);
        dummy.next=head;

        ListNode *pre=head,*cur=head->next;
        while(cur){
            pre->next=cur->next;
            cur->next=dummy.next;
            dummy.next=cur;
            cur=pre->next;
        }
        return dummy.next;
    }
};

全程pre指向头结点，其之后的节点为该轮要进行头插的节点。

以下
https://leetcode-cn.com/problems/reverse-linked-list/solution/c-die-dai-fa-di-gui-fa-by-eric-345/

迭代法

和头插法大同小异，看上去更简单。

ListNode* reverseList(ListNode* head) {

    if (head == nullptr || head->next == nullptr) 
        return head;

    ListNode* prev = nullptr;
    ListNode* cur = head;

    while (cur)
    {
        ListNode* tmp = cur->next; //保存下轮要前插的节点
        cur->next = prev; //将cur节点插到新链表的头结点前
        prev = cur;//这时头结点变为本轮前移的节点
        cur = tmp;//之前保存的节点变为当前节点
    }

    return prev;    
}

迭代法

画图从后向前分析吧。

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(!head||!head->next) return head;
        ListNode *p=reverseList(head->next);

        head->next->next=head;
        head->next=nullptr;

        return p;        
    }
};

附以下playgroud单句调试，方便理解递归的细节，容许我懒一下，就不一句一句分析了。

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(!head||!head->next) return head;
        ListNode *p=reverseList(head->next);
        
        string out = listNodeToString(head);
        cout <<"@1 "<<"head "<< out << endl;
        
        out = listNodeToString(p);
        cout <<"p "<< out << endl;

        head->next->next=head;
        //out = listNodeToString(head);
        //cout <<"@2 "<<"head "<< out << endl;
        head->next=nullptr;
        out = listNodeToString(head);
        cout <<"@3 "<<"head "<< out << endl;
        
        out = listNodeToString(p);
        cout <<"p "<< out << endl;

        return p;        
    }

@1 head [4, 5]
p [5]
@3 head [4]
p [5, 4]
@1 head [3, 4]
p [5, 4]
@3 head [3]
p [5, 4, 3]
@1 head [2, 3]
p [5, 4, 3]
@3 head [2]
p [5, 4, 3, 2]
@1 head [1, 2]
p [5, 4, 3, 2]
@3 head [1]
p [5, 4, 3, 2, 1]
[5, 4, 3, 2, 1]