链表的反转问题(Reverse Linked List)

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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;    }};

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