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.
部分反转链表，细心点处理好各种情况就好
var reverseBetween = function(head, m, n) {
if (m === n || !head)
return head;
var step = m-1;
var pre = null;
//将pre指向要反转的前一个
while (step>0) {
step--;
pre = pre ? pre.next : head;
}
step = n - m;
//不反转的末尾
var ne = pre;
//反转的开头
var rb = pre ? pre.next : head;
pre = rb;
//要反转的
var now = pre.next;
while (step>=1) {
var temp = now.next;
now.next = pre;
pre = now;
now = temp;
step--;
}
//反转的开头指向后面
rb.next = now;
if (ne) {
//如果ne不是空的，意味着不是从头开始反转的
//ne指向被反转的链表的末尾
ne.next = pre;
return head;
}
//ne是空的，那就意味着整个链表的头是反转链表的末尾
return pre;
};
