本文整理汇总了C#中ListNode类的典型用法代码示例。如果您正苦于以下问题：C# ListNode类的具体用法？C# ListNode怎么用？C# ListNode使用的例子？那么恭喜您, 这里精选的类代码示例或许可以为您提供帮助。
ListNode类属于命名空间，在下文中一共展示了ListNode类的15个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: RemoveNthFromEnd
public ListNode<int> RemoveNthFromEnd(ListNode<int> head, int n)
{
// assume the input is always valid
if (head.Next == null) // only 1 node
{
return null;
}
// reach that node first
var tail = head;
var preDelete = head;
for (int i = 0; i < n; i++)
{
tail = tail.Next;
}
if (tail == null) // toDelete is head
{
return head.Next;
}
while (tail.Next != null)
{
tail = tail.Next;
preDelete = preDelete.Next;
}
preDelete.Next = preDelete.Next.Next;
return head;
}示例2: ReverseSingleSublistTest
public void ReverseSingleSublistTest()
{
Func<ListNode<int>, int, int, ListNode<int>>[] functions = new Func<ListNode<int>, int, int, ListNode<int>>[]
{
ReverseSingleSublist.BruteForce,
ReverseSingleSublist.SinglePass
};
for(int i = 0; i < 10; i++)
{
int[] data = ArrayUtilities.CreateRandomArray(10, 0, 10);
for (int start = 1; start <= 10; start++)
{
for(int end = start; end <= 10; end++)
{
ListNode<int>[] results = new ListNode<int>[functions.Length];
for (int j = 0; j < results.Length; j++)
{
ListNode<int> head = LinkedListUtilities.Initialize(data);
results[j] = functions[j](head, start, end);
Assert.IsTrue(LinkedListUtilities.AreEqual(results[0], results[j]));
}
}
}
}
}示例3: AddTwoNumbers
public ListNode AddTwoNumbers(ListNode l1, ListNode l2)
{
var add1=Convert(l1);
var add2=Convert(l2);
var res=add1+add2;
return Convert(res);
}示例4: MergeTwoLists
//method to merge two sorted lists
private ListNode MergeTwoLists(ListNode l1, ListNode l2){
if(l1==null||l2==null){
if(l1 ==null) return l2;
if(l2 ==null) return l1;
}
ListNode newHead,p,p1,p2;
if(l1.val < l2.val){
newHead = l1;
p1 = l1.next;
p2 = l2;
}
else{
newHead = l2;
p1 = l1;
p2 = l2.next;
}
p = newHead;
while(p1!=null && p2!=null){
if(p1.val < p2.val){
p.next = p1;
p1 = p1.next;
}
else{
p.next = p2;
p2 = p2.next;
}
p=p.next;
}
//no need to use the while loop, directly append p1 or p2 to p
//because rest nodes of p1 or p2 are already sorted
if(p1!=null) p.next=p1;
if(p2!=null) p.next=p2;
return newHead;
}示例5: ReverseBetween
public ListNode ReverseBetween(ListNode head, int m, int n)
{
var length = n-m +1;
if(m == 1){
ListNode tail;
var h = ReverseList(head, length, out tail);
head.next = tail;
return h;
}
else{
var c = 2;
var t = head;
while(c != m){
t = t.next;
c++;
}
ListNode tail;
var h = ReverseList(t.next, length, out tail);
t.next.next = tail;
t.next = h;
return head;
}
}示例6: ReverseBetween
public ListNode ReverseBetween(ListNode head, int m, int n) {
if(head==null) return null; //handle edge case
ListNode preHead = new ListNode(0);
preHead.next = head;
ListNode start = preHead; //start will be the (m-1)th node in the list
//advance start pointer to the (m-1)th node in the list
int count = 1;
while(start!=null && count<m){
start = start.next;
count++;
}
if(count<m) return head; //length of the list is smaller than m, edge case
//end is the mth node in the list
ListNode end = start.next; //node will be inserted between start and end
ListNode curr = end.next;//curr is the node to be insert
//insert each node right after start, advance curr until count==n
while(curr!=null && count<n){
ListNode next = curr.next; //get the next node using curr.next
curr.next = start.next; //insert curr right after start node
start.next = curr;
end.next = next; //point end to next node
curr = next; //advance curr to next node
count++;
}
return preHead.next; //preHead is the node before original head, so return preHead.next
}示例7: Print
private static void Print(ListNode n)
{
if(n != null){
Console.Write(n.val);
Print(n.next);
}
}示例8: SwapPairs
public ListNode<int> SwapPairs(ListNode<int> head)
{
if (head == null || head.Next == null)
{
return head;
}
ListNode<int> preResult = new ListNode<int>(-9999); // helper node
var pre = preResult;
var n1 = head;
var n2 = head.Next;
var post = n2.Next;
while (true)
{
pre.Next = n2;
n2.Next = n1;
n1.Next = post;
if (post == null || post.Next == null)
{
break;
}
pre = n1;
n1 = post;
n2 = n1.Next;
post = n2.Next;
}
return preResult.Next;
}示例9: ReverseBetween
public ListNode<int> ReverseBetween(ListNode<int> head, int m, int n)
{
if (m == n)
{
return head;
}
// find pre-changing head
ListNode<int> dummy = new ListNode<int>(0);
dummy.Next = head;
var preChaningHead = dummy;
int changeLength = n - m;
while (m-- > 1)
{
preChaningHead = preChaningHead.Next;
}
var curNode = preChaningHead.Next;
for (int i = 0; i < changeLength; i++)
{
var preNext = preChaningHead.Next;
var curNext = curNode.Next;
curNode.Next = curNext.Next;
preChaningHead.Next = curNext;
curNext.Next = preNext;
}
return dummy.Next;
}示例10: AddTwoNumbers
public ListNode<int> AddTwoNumbers(ListNode<int> l1, ListNode<int> l2)
{
ListNode<int> current = new ListNode<int>(-1);
ListNode<int> head = current;
int carry = 0;
while (l1 != null || l2 != null || carry > 0)
{
int result = carry;
if (l1 != null)
{
result += l1.Val;
l1 = l1.Next;
}
if (l2 != null)
{
result += l2.Val;
l2 = l2.Next;
}
current.Next = new ListNode<int>(result % 10);
current = current.Next;
carry = result / 10;
}
return head.Next;
}示例11: Partition
public ListNode<int> Partition(ListNode<int> head, int x)
{
if (head == null)
return null;
ListNode<int> first = new ListNode<int>(-1);
ListNode<int> second = new ListNode<int>(-1);
ListNode<int> originalFirst = first;
ListNode<int> originalSecond = second;
while (head != null)
{
if (head.Val < x)
{
first.Next = head;
first = first.Next;
}
else
{
second.Next = head;
second = second.Next;
}
head = head.Next;
}
second.Next = null;
first.Next = originalSecond.Next;
return originalFirst.Next;
}示例12: AddTwoNumbers
public ListNode<int> AddTwoNumbers(ListNode<int> l1, ListNode<int> l2)
{
// add l2 on to l1
ListNode<int> current = new ListNode<int>(0);
var answer = current;
int toAdd = 0;
while (true)
{
current.Val += toAdd;
if (l1 != null)
{
current.Val += l1.Val;
l1 = l1.Next;
}
if (l2 != null)
{
current.Val += l2.Val;
l2 = l2.Next;
}
toAdd = current.Val / 10;
current.Val = current.Val % 10;
if (l1 != null || l2 != null || toAdd > 0)
{
current.Next = new ListNode<int>(0);
current = current.Next;
}
else
{
break;
}
}
return answer;
}示例13: Process
public static ListNode Process(ListNode head, uint k)
{
if (head == null || k == 0)
{
return null;
}
ListNode ahead = head;
for (int i = 0; i < k - 1; ++i)
{
if (ahead.Next != null)
{
ahead = ahead.Next;
}
else
{
return null;
}
}
ListNode behind = head;
while (ahead.Next != null)
{
ahead = ahead.Next;
behind = behind.Next;
}
return behind;
}示例14: MergeKLists1
public ListNode MergeKLists1(ListNode[] lists)
{
ListNode result = new ListNode(0);
ListNode pointer = result;
MinHeap mh = new MinHeap();
foreach (ListNode node in lists)
{
if (node != null)
{
mh.Add(node);
}
}
while (mh.Count > 0)
{
ListNode node = mh.PopMin();
if (node.next != null)
{
mh.Add(node.next);
}
pointer.next = node;
pointer = pointer.next;
}
return result.next;
}示例15: Main
public static void Main()
{
//test#1: null
ListNode head = null;
Solution.display(Solution.removeNodes(head,5)); //empty
//test#2: 1
head = new ListNode(1);
Solution.display(Solution.removeNodes(head,5)); //1->
//test#3: 5
head = new ListNode(5);
Solution.display(Solution.removeNodes(head,5)); //empty
//test#4: 1->2
head = new ListNode(1);
head.next = new ListNode(2);
Solution.display(Solution.removeNodes(head,5)); //1->2->
//test#5: 5->5->5
head = new ListNode(5);
head.next = new ListNode(5);
head.next.next = new ListNode(5);
Solution.display(Solution.removeNodes(head,5)); //empty
//test#6: 1->2->5->5
head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(5);
head.next.next.next = new ListNode(5);
Solution.display(Solution.removeNodes(head,5)); //1->2->
//test#7: 1->2->5->5->3
head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(5);
head.next.next.next = new ListNode(5);
head.next.next.next.next = new ListNode(3);
Solution.display(Solution.removeNodes(head,5)); //1->2->3->
}