206. Reverse (Linked List)

题目：

Reverse a singly linked list.
Hint:
A linked list can be reversed either iteratively or recursively. Could you implement both?

iteratively

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* reverseList(struct ListNode* head) {    if(head == NULL) return;    struct ListNode *pre, *cur, *next;    pre = NULL;    cur = head;    while(cur){//就地        next = cur->next;        cur->next = pre;        pre = cur;        cur = next;    }    return pre;}

runtime:4ms

recursively

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */ // 1->2->3struct ListNode* reverseList(struct ListNode* head) {    if(head==NULL){        return head;    }    struct ListNode *p, *q;    p = head;    q = head->next;    if(q == NULL){        return head;    }    else{        head = reverseList(q);//递归到最深处始终为3    }    q->next = p;    p->next = NULL;    return head;}

runtime:4ms

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* reverseList(struct ListNode* head) {    struct ListNode *p;    p = head;    // q = head->next;    if(head==NULL || head->next == NULL){//空或最后结点        return head;    }    head = reverseList(p->next);    p->next->next = p;    p->next = NULL;    return head;}

runtime:0ms

递归分析

递归算法：
- 若只有一个结点，则直接返回;
- 若存在两个结点（a1,a2）则需要做的操作有：
a2->next=a1;a1->next=NULL;return a2;//a2即新的头结点
- 若有三个结点,则应先将子链（a2,a3）先逆置且返回该子链的新的头结点，然后把子链（a2,a3）当作一个复合结点a2’ //组成新的二元组（a1,a2’）
然后就可以执行前面相同的操作：
a2'->next=a1;a1->next=NULL;return a3';即可;
- 多个结点同理可得。
已经阐述过，我很水，所以就各种参考。
感谢这个和这个。