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Problem:
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.
分析:
链表的反转是经典的问题,Reorder List中问题中也有涉及。三个指针,一个维护头部元素的指针(head),一个维护新链表尾部的指针(tail),另外一个是将要插入到链表头部元素的指针(cur)。顺序从第二个元素开始,依次向后取每个元素,将其插入链表的头部。时间复杂度O(n)。
1 if(head == NULL) 2 return; 3 4 ListNode *tail = head; 5 while(tail->next) { 6 ListNode *cur = tail->next; 7 tail->next = cur->next; 8 cur->next = head; 9 head = cur; 10 }
这个题目唯一不同点在于,需要获取到第m个节点的前驱,使链表能够接好。为了代码的简洁与一致,避免判断m=1为头节点时的麻烦,可以在头部加入哨兵(代码第7-8行),然后循环找到m节点的前驱(第9-12行),而后的反转就是经典的问题了(第13-19行)。
1 class Solution { 2 public: 3 ListNode *reverseBetween(ListNode *head, int m, int n) { 4 if(head == NULL || m == n || m < 1 || m > n) 5 return head; 6 7 ListNode *sentinel = new ListNode(0); 8 sentinel->next = head; 9 ListNode *pre = sentinel; 10 for(int i = 1; i < m; ++i) { 11 pre = pre->next; 12 } 13 ListNode *tail = pre->next; 14 for(int i = m; i < n; ++i) { 15 ListNode *cur = tail->next; 16 tail->next = cur->next; 17 cur->next = pre->next; 18 pre->next = cur; 19 } 20 head = sentinel->next; 21 delete sentinel; 22 return head; 23 } 24 };
Reverse Linked List II,码迷,mamicode.com
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原文地址:http://www.cnblogs.com/foreverinside/p/3702128.html