'lintcode 翻转链表2'

描述

翻转链表中第m个节点到第n个节点的部分

m，n满足1 ≤ m ≤ n ≤ 链表长度

样例

例1:

输入: 1->2->3->4->5->NULL, m = 2 and n = 4,
输出: 1->4->3->2->5->NULL.
例2:

输入: 1->2->3->4->NULL, m = 2 and n = 3,
输出: 1->3->2->4->NULL.

挑战

Reverse it in-place and in one-pass

思考

m到n这段的翻转直接使用翻转链表的算法即可，只需要把循环的条件改为n-m+1即可。主要的是解决翻转链表的前后2个如何处理。这里设置了两个变量 tempFirst 和 tempLast 分别来表示第m-1个 和翻转完成之后的第n-1个。这样处理的问题是m=1的时候要重新处理。

代码

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/** * Definition of singly-linked-list: * class ListNode { * public: * int val; * ListNode *next; * ListNode(int val) { * this->val = val; * this->next = NULL; * } * } */ class Solution { public: /** * @param head: ListNode head is the head of the linked list * @param m: An integer * @param n: An integer * @return: The head of the reversed ListNode */ ListNode * reverseBetween(ListNode * head, int m, int n) { // write your code here if (!head || !head->next || n == m) return head; int delt = n-m+1; ListNode *cur = head, *prev = NULL, *tempLast = NULL, *tempFirst = head, *nex = NULL; int M = m; if (M > 1){ while (m-2) { cur = cur->next; tempFirst = tempFirst->next; m--; } cur = cur->next; } tempLast = cur; while (delt) { nex = cur->next; cur->next = prev; prev = cur; cur = nex; delt--; } if (M == 1) { tempFirst->next = cur; return prev; } tempFirst->next = prev; tempLast->next = cur; return head; } };

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