[LeetCode92]Reverse Linked List II

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题目：

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.

反转m到n处的单链表

The basic idea is as follows:

(1) Create a new_head that points to head and use it to locate the immediate node before them-th (notice that it is 1-indexed) node pre;

(2) Set cur to be the immediate node after pre and at each time move the immediate node after cur (named move) to be the immediate node after pre. Repeat it for n - m times.

分类：Linked List

代码：

class Solution {  
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        ListNode* new_head = new ListNode(0);
        new_head -> next = head;
        ListNode* pre = new_head;
        for (int i = 0; i < m - 1; i++)
            pre = pre -> next;
        ListNode* cur = pre -> next;
        for (int i = 0; i < n - m; i++) {
            ListNode* move = cur -> next; 
            cur -> next = move -> next;
            move -> next = pre -> next;
            pre -> next = move;
        }
        return new_head -> next;
    }
}; 

[LeetCode92]Reverse Linked List II

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原文地址：http://www.cnblogs.com/zhangbaochong/p/5287562.html