Linked list notes

如何判断while循环里面的条件。一般来说是结束的时候，如果不清楚则看while里面，需要用到next的ListNode不为空。

如果node在后面的操作中会丢失，那么先要cache。比如reverse里面的，cur.next会丢失。
1. Reverse
其实reserse就只有有个操作，cur.next = prev，其它的就是调整pre 和cur。

     ListNode prev = null;  
     ListNode cur = mid;  
     while(cur != null) {  
       ListNode temp = cur.next;  
       cur.next = prev;  
       prev = cur;  
       cur = temp;  
     }  
     return prev;  

2. Merge

  while (l1 != null) {  
     ListNode n1 = l1.next, n2 = l2.next;  
     l1.next = l2;  
     if (n1 == null)  
      break;  
     l2.next = n1;  
     l1 = n1;  
     l2 = n2;  
    }  

3. Find mid and cut

 ListNode slow = head;  
     ListNode fast = head;  
     ListNode prev = null;  
     while(fast != null && fast.next != null) {  
       prev = slow;  
       slow = slow.next;  
       fast = fast.next.next;  
     }  
     prev.next =null;  
     return slow;  

4. Determine cycle

 public boolean hasCycle(ListNode head) {  
   if (head == null || head.next == null) {  
     return false;  
   }  
   ListNode slow = head;  
   ListNode fast = head.next;  
   while (slow != fast) {  
     if (fast == null || fast.next == null) {  
       return false;  
     }  
     slow = slow.next;  
     fast = fast.next.next;  
   }  
   return true;  
 }  

5. Merge two sorted list

 /**  
  * Definition for singly-linked list.  
  * public class ListNode {  
  *   int val;  
  *   ListNode next;  
  *   ListNode(int x) { val = x; }  
  * }  
  */  
 class Solution {  
   public ListNode mergeTwoLists(ListNode l1, ListNode l2) {  
     ListNode dummy = new ListNode(0);  
     ListNode res = new ListNode(0);  
     dummy = res;  
     while (l1 != null && l2 != null) {  
       if (l1.val < l2.val) {  
         res.next = l1;  
         l1 = l1.next;  
       } else {  
         res.next = l2;  
         l2 = l2.next;  
       }  
       res = res.next;  
     }  
     while (l1 !=null) {  
       res.next = l1;  
     }  
     if(l2 != null) {  
       res.next = l2;;  
     }  
     return dummy.next;  
   }  
 }