首页 > 代码库 > 14. Reverse Linked List II
14. Reverse Linked List II
Reverse Linked List II
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.
Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list.
总结：其实就是反转链表。不过是反转中间一部分。要注意的是保存第一个结点的前继的指针； 若第一个结点是头结点，注意反转子串的尾结点变为头结点。
/** * 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 *preNode1, *node1, *node2; preNode1 = node1 = node2 = NULL; int cnt = 0; for(ListNode *p = head; p != NULL; p = p->next) { cnt++; if(cnt == m-1) preNode1 = p; // hidden: m > 1 if(cnt == m) node1 = p; if(cnt == n) { node2 = p; break; } } ListNode *tail = node2->next; // must take out as a tag. ListNode *pre = tail, *post; while(node1 != tail) { post = node1->next; node1->next = pre; pre = node1; node1 = post; } if(m > 1) { preNode1->next = node2; return head;} return node2; // node1 is the 1st node. } };
14. Reverse Linked List II
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。