链表反转

List

pre

3

1

prev和cur两个指针循环移动

5

4

next

prev

Node *List

while(cur) {   next = cur->next;//1   cur->next = pre;  //2   pre = cur;           //3   cur = next;         //4}

loop

cur

2

next确保链表不断

X

Node *pre = NULL;Node *cur = List;Node *next = NULL;