标签：style blog os io strong for div log amp
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.
总结：其实就是反转链表。不过是反转中间一部分。要注意的是保存第一个结点的前继的指针； 若第一个结点是头结点，注意反转子串的尾结点变为头结点。
/**
* 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 *preNode1, *node1, *node2;
preNode1 = node1 = node2 = NULL;
int cnt = 0;
for(ListNode *p = head; p != NULL; p = p->next) {
cnt++;
if(cnt == m-1) preNode1 = p; // hidden: m > 1
if(cnt == m) node1 = p;
if(cnt == n) { node2 = p; break; }
}
ListNode *tail = node2->next; // must take out as a tag.
ListNode *pre = tail, *post;
while(node1 != tail) {
post = node1->next;
node1->next = pre;
pre = node1;
node1 = post;
}
if(m > 1) { preNode1->next = node2; return head;}
return node2; // node1 is the 1st node.
}
};
标签：style blog os io strong for div log amp
原文地址：http://www.cnblogs.com/liyangguang1988/p/3933965.html
