链表的反转问题(Reverse Linked List)

反转单链表思路有三种，

1.将1->2->3变为 2->1->3再变为3->2->1（头插法）

2.将1->2->3变成1<-2->3再变为1<-2-<3（反转法）

3.递归

下面我们分别进行如下编码

1.头插法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head == NULL || head->next == NULL) return head;
        ListNode prehead(-1);
        prehead.next = head;
        ListNode *pre = &prehead;
        ListNode *cur = head;

        while(cur&&cur->next) {
            ListNode *tmp = pre->next;
            pre->next = cur->next;
            cur->next = cur->next->next;
            pre->next->next = tmp;
        }
        return prehead.next;   
    }
};

2.反转法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode *pre = NULL;
        while(head){
            ListNode* tmp = head->next; 
            head->next = pre;
            pre = head;
            head = tmp;
        }
        return pre;
    }
};

3.递归法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(!head || !head->next) return head;
        ListNode* node = reverseList(head->next);
        head->next->next = head;
        head->next = NULL;
        return node;
    }
};

有了最基本的链表反转操作，我们可以解决一些有意思的问题

比如判断链表是否有回文

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if(head == NULL || head->next == NULL) return true;
        ListNode* slow = head;
        ListNode* fast = head;
        ListNode* p = head;

        while(fast->next != NULL && fast->next->next !=NULL) {
            slow = slow->next;
            fast = fast->next->next;
        }
        slow->next = reverseList(slow->next);
        slow = slow->next;
        while(slow){
            if(head->val != slow->val)
                return false;
            head = head->next;
            slow = slow->next;
        }
        return true;
    }

    ListNode* reverseList(ListNode* head){
        ListNode* pre =NULL;
        ListNode* next = NULL;
        while(head != NULL) {
            next = head->next;
            head->next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
};