在 C# 中,有多种集合类可供使用,它们分别适用于不同的场景,部分代码示例提供了LeetCode相关的代码应用。
int[] numbers = new int[5] { 1, 5, 2, 3, 4 };
numbers[0] = 1;//update
int firstNumber = numbers[0];
int lastNumber = numbers[numbers.Length - 1];
Array.Sort(numbers);//排序
LeetCode: 106. 从中序与后序遍历序列构造二叉树 - 力扣(LeetCode)
public TreeNode BuildTree(int[] inorder, int[] postorder)
{
if (inorder.Length == 0 || postorder.Length == null) return null;
int rootValue = postorder.Last();
TreeNode root = new TreeNode(rootValue);
int delimiterIndex = Array.IndexOf(inorder, rootValue);
root.left = BuildTree(inorder.Take(delimiterIndex).ToArray(), postorder.Take(delimiterIndex).ToArray());
root.right = BuildTree(inorder.Skip(delimiterIndex + 1).ToArray(), postorder.Skip(delimiterIndex).Take(inorder.Length - delimiterIndex - 1).ToArray());
return root;
}
List numberList = new List();
numberList.Add(1);
int firstElement = numberList[0];
LinkedList numberLinkedList = new LinkedList();
numberLinkedList.AddLast(1);
numberLinkedList.AddFirst(2);
numberLinkedList.AddFirst(3);
numberLinkedList.Remove(1);
numberLinkedList.RemoveFirst();
numberLinkedList.RemoveLast();
int count=numberLinkedList.Count;
bool isContains= numberLinkedList.Contains(3);
Push
)和出栈(Pop
)操作的时间复杂度为O(1) 。 Stack numberStack = new Stack();
numberStack.Push(1);
// Removes and returns the object at the top of the System.Collections.Generic.Stack`1.
int topElement = numberStack.Pop();
// Returns the object at the top of the System.Collections.Generic.Stack`1 without removing it.
topElement = numberStack.Peek();
int count = numberStack.Count;
LeetCode: 144. 二叉树的前序遍历
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public IList PreorderTraversal(TreeNode root) {
var result = new List();
if (root == null) return result;
Stack treeNodes = new Stack();// 栈是先进后出。 中序遍历是:中--》 左--》右
treeNodes.Push(root);
while (treeNodes.Count > 0)
{
TreeNode current = treeNodes.Pop();
result.Add(current.val);// 中
if (current.right != null)// 右
{
treeNodes.Push(current.right);
}
if (current.left != null)// 左
{
treeNodes.Push(current.left);
}
}
return result;
}
}
Enqueue
)和出队(Dequeue
)操作的时间复杂度为 O(1)。 Queue numberQueue = new Queue();
numberQueue.Enqueue(1);
// Removes and returns the object at the beginning of the System.Collections.Generic.Queue`1.
int firstElement = numberQueue.Dequeue();
//Returns the object at the beginning of the System.Collections.Generic.Queue`1 without removing it.
numberQueue.Peek();
int count = numberQueue.Count;
LeetCode: 102. 二叉树的层序遍历 - 力扣(LeetCode)
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public IList> LevelOrder(TreeNode root) {
var result = new List>();
Queue queue = new ();
if (root != null) queue.Enqueue(root);
while (queue.Count>0)
{
var row=new List();
int count=queue.Count;// 需要事先记录Queue的Count,不能直接依靠Queue本身
for (int i=0; i < count; i++)
{
var currentNode=queue.Dequeue();
row.Add(currentNode.val);
if (currentNode.left != null) queue.Enqueue(currentNode.left);
if (currentNode.right != null) queue.Enqueue(currentNode.right);
}
result.Add(row);
}
return result;
}
}
HashSet numberHashSet = new HashSet();
numberHashSet.Add(1);
bool isAddSuccess = numberHashSet.Add(1);// false
bool containsOne = numberHashSet.Contains(1);// true
如LeetCode:491. 非递减子序列
public class Solution {
public IList> FindSubsequences(int[] nums) {
var result = new List>();
var path = new List();
if(nums==null || nums.Length == 0) return result;
BackTacking(nums,result,path,0);
return result;
}
private void BackTacking(int[] nums,List> result, List path,int startIndex)
{
if (path.Count>=2) result.Add(new List(path));
HashSet used =new HashSet();
for (int i=startIndex;i0 && nums[i]
Dictionary scores = new Dictionary();
scores.Add("Alice", 90);
int aliceScore = scores["Alice"];
LeetCode: 105. 从前序与中序遍历序列构造二叉树 - 力扣(LeetCode)
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public TreeNode BuildTree(int[] preorder, int[] inorder)
{
if (preorder == null || inorder == null)
{
return null;
}
/*
前序遍历 根 左子树 右子树
中序遍历 左子树 根 右子树
*/
if (preorder.Length != inorder.Length)
{
return null;
}
Dictionary keyMap = new Dictionary();
for (int i = 0; i < inorder.Length; i++)
{
keyMap.Add(inorder[i], i);
}
return BuildTree(preorder: preorder, keyMap, preleft: 0, preRight: preorder.Length - 1, inLeft: 0, inRight: inorder.Length - 1);
}
///
///
///
///
/// 用于确定pIndex。Root的Index
///
///
///
///
///
private TreeNode BuildTree(int[] preorder, Dictionary keyMap, int preleft, int preRight, int inLeft, int inRight)
{
if (preleft > preRight || inLeft > inRight)
{
return null;
}
int rootValue = preorder[preleft];
TreeNode treeNode = new TreeNode(rootValue);
int pIndex = keyMap[rootValue];
treeNode.left = BuildTree(preorder, keyMap, preleft: preleft + 1, preRight: pIndex + preleft - inLeft, inLeft, inRight: pIndex - 1);//根据PreOrder去取Left node
treeNode.right = BuildTree(preorder, keyMap, preleft: pIndex + preleft - inLeft + 1, preRight: preRight, inLeft: pIndex + 1, inRight);//根据inOrder去取Right node
return treeNode;
}
}
SortedSet sortedNumbers = new SortedSet();
sortedNumbers.Add(3);
sortedNumbers.Add(1);
// 元素会自动排序,遍历输出为 1, 3
foreach (int number in sortedNumbers)
{
Console.WriteLine(number);
}
SortedDictionary sortedScores = new SortedDictionary();
sortedScores.Add("Evan", 38);
sortedScores.Add("Alice", 30);
// 会根据键排序,遍历输出 Alice: 30, Evan: 38
foreach (KeyValuePair pair in sortedScores)
{
Console.WriteLine($"{pair.Key}: {pair.Value}");
}