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.
用了栈。注意指针移动时不同指针的分工。
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/** * 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 *guard = new
ListNode(0); guard -> next =head; ListNode *pre = guard; ListNode *temp = head; ListNode *re ; stack<ListNode*> sk; for(int
i =1 ; i < m ;i++) { pre = pre->next; temp = pre ->next; } ListNode *tail = temp; temp =temp->next; for(int
i = m ; i < n ; i++) { sk.push(temp); temp =temp->next; } while(!sk.empty()) { ListNode* tt =sk.top(); pre->next = tt; pre =pre->next; sk.pop(); } pre->next = tail; tail->next =temp; return
guard->next; }}; |
Reverse Linked List II,布布扣,bubuko.com
原文：http://www.cnblogs.com/pengyu2003/p/3605189.html