206&92 反转链表

206.反转链表

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

代码：

/**
 * 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 *p = head;//p = 1
        ListNode *after;
        ListNode *temp;
        if (p == NULL)
          return p;
        after = p->next;//after = 2
        p->next = NULL;//1->NULL
        while(after!=NULL)
        {
            temp = after->next;//temp = 3
            after->next = p;//2->1  3->2
            p = after;//p = 2  p = 3
            after = temp;//after = 3 after = NULL
        }
        return p;
    }
};

92 反转链表2

难度中等434反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。

说明:
1 ≤ m ≤ n ≤ 链表长度。

示例:

输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL

 

 

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode *p, *after, *temp, *end, *before;
        p = head;
        before = head;
        if(p == NULL)
        return p;
        int i;
        for(i = 1; i <= m-1; i++)
        {
            before = p;
            p = p->next;
        }
        end = p;//p = a[m]
        after = p->next;//after = a[m+1] 
        for(i = m+1; i <= n; i++)
        {
            temp = after->next;//temp = a[m+2] temp = a[m+3] temp = NULL
            after->next = p;//a[m+1]->a[m]  a[m+2]->a[m+1] a[m+3]->a[m+2]
            p = after;//p = a[m+1] p = a[m+2] p = a[m+3]
            after = temp; //after = a[m+2] after = a[m+3] after = NULL
        }
        end->next = after;
        if(m != 1)
        {
          before->next = p;
          return head;
        }
        else
           return p;
    }
};