Leetcode[92]-Reverse Linked List II
2015-06-10 22:41
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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.
思路:分别找到m节点和m的前节点,n节点和n的后节点,然后翻转m到n的部分,最后链接三部分成一个整体;代码如下
Code(c++):
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.
思路:分别找到m节点和m的前节点,n节点和n的后节点,然后翻转m到n的部分,最后链接三部分成一个整体;代码如下
Code(c++):
/** * 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) { if(head == NULL) return head; ListNode *newList = new ListNode(-1); newList->next = head; ListNode *prebegin = newList; ListNode *begin = head; ListNode *end = newList; ListNode *postend = head; //begin as center start node while(--m){ prebegin = prebegin->next; begin = begin->next; } //end as center end node while(n--){ end = end->next; postend = postend->next; } //reverse center part reverse(begin, end); //link three part as one list prebegin->next = begin; end->next = postend; head = newList->next; return head; } void reverse(ListNode*& begin, ListNode*& end){ if(begin == end) return; if(begin->next == end){ end->next = begin; } else { //at least 3 nodes ListNode* pre = begin; ListNode* cur = pre->next; ListNode* post = cur->next; while(post != end->next){ cur->next = pre; pre = cur; cur = post; post = post->next; } cur->next = pre; } //swap begin and end ListNode* temp = begin; begin = end; end = temp; } };
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