92. Reverse Linked List II
摘要：
Reverse a linked list from position m to n. Do it in one-pass.
Note: 1 ≤ m ≤ n ≤ length of list.
Example:
Input: 1->2->3->4-&...
Reverse a linked list from position m to n. Do it in one-pass.
Note: 1 ≤ m ≤ n ≤ length of list.
Example:
Input: 1->2->3->4->5->NULL, m = 2, n = 4 Output: 1->4->3->2->5->NULL
難度：medium
題目：反轉從m到n的連結串列元素。一次遍歷。
思路：記錄m及m之前的位置，然後使用頭插法。
Runtime: 2 ms, faster than 97.09% of Java online submissions for Reverse Linked List II.
Memory Usage: 36.9 MB, less than 0.95% of Java online submissions for Reverse Linked List II.
/**
* Definition for singly-linked list.
* public class ListNode {
*int val;
*ListNode next;
*ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
if (m == n) {
return head;
}
ListNode dummyHead = new ListNode(0);
dummyHead.next = head;
ListNode ptr = head, prevMPtr = dummyHead, tailPtr = head;
for (int i = 1; i <= n; i++) {
ListNode node = ptr;
ptr = ptr.next;
if (i == m - 1) {
prevMPtr = node;
} else if (i == m) {
tailPtr = node;
node.next = null;
} else if (i > m) {
node.next = prevMPtr.next;
prevMPtr.next = node;
}
}
tailPtr.next = ptr;
return dummyHead.next;
}
}