コード例 #1
0
 /**
  * The basic idea is to have two lists: one for nodes smaller than x; the other for nodes greater
  * or equal to x. We delete nodes one by one from the orignal list, adding each node to either of
  * the two lists. The logic of this solution is simpler than only using one extra list, and thus
  * less error-prone.
  */
 public ListNode partition(ListNode head, int x) {
   if (head == null || head.next == null) {
     return head;
   }
   ListNode dummy = new ListNode(0);
   dummy.next = head;
   ListNode l1 = new ListNode(0);
   ListNode t1 = l1;
   ListNode l2 = new ListNode(0);
   ListNode t2 = l2;
   while (dummy.next != null) {
     ListNode tmp = dummy.next;
     dummy.next = dummy.next.next;
     if (tmp.val < x) {
       t1.next = tmp;
       t1 = t1.next;
     } else {
       t2.next = tmp;
       t2 = t2.next;
     }
   }
   t1.next = l2.next;
   // always remember to set the tail's next pointer to null! Another way is to set the next
   // pointer of a deleted node
   // to null as soon as we delete it.
   t2.next = null;
   return l1.next;
 }
コード例 #2
0
 public ListNode partition1(ListNode head, int x) {
   if (head == null || head.next == null) {
     return head;
   }
   ListNode l1 = new ListNode(0);
   l1.next = head;
   ListNode l2 = new ListNode(0);
   ListNode tail = l2;
   ListNode p = l1;
   while (p.next != null) {
     if (p.next.val < x) {
       ListNode tmp = p.next;
       p.next = p.next.next;
       tail.next = tmp;
       tail = tail.next;
     } else {
       p = p.next;
     }
   }
   tail.next = l1.next;
   return l2.next;
 }