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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.
大清早起来就被链表虐哭了啊， 看了下别人的，额原来可以这么简单，果然脑子还是转不过来的，实际上是很常见的一个题目，代码很简单，完全不用注释也可以的：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* reverseBetween(ListNode* head, int m, int n) { 12 if(head == NULL) return NULL; 13 ListNode * p = head; 14 int i, j; 15 for(i = 1; i < m; ++i){ 16 p = p->next; 17 } 18 ListNode * q = p; 19 for(i = m; i < n; ++i){ 20 for(j = i; j < n; ++j){ 21 q = q->next; 22 } 23 swap(p->val, q->val); 24 n--; 25 p = p->next; 26 q = p; 27 } 28 return head; 29 } 30 };
注意一下那个swap， swap用的很巧妙。
之后有看到一个大神写出来的，也很简单，贴出来学习一个：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *reverseBetween(ListNode *head, int m, int n) { 12 // Start typing your C/C++ solution below 13 // DO NOT write int main() function 14 if (head == NULL) 15 return NULL; 16 17 ListNode *q = NULL; 18 ListNode *p = head; 19 for(int i = 0; i < m - 1; i++) 20 { 21 q = p; 22 p = p->next; 23 } 24 25 ListNode *end = p; 26 ListNode *pPre = p; 27 p = p->next; 28 for(int i = m + 1; i <= n; i++) 29 { 30 ListNode *pNext = p->next; 31 32 p->next = pPre; 33 pPre = p; 34 p = pNext; 35 } 36 37 end->next = p; 38 if (q) 39 q->next = pPre; 40 else 41 head = pPre; 42 43 return head; 44 } 45 };
唉唉，经常遇到链表脑子就转不过来，这个还是要多练练啊。
LeetCode OJ：Reverse Linked List II（反转链表II）
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原文地址：http://www.cnblogs.com/-wang-cheng/p/4874091.html
