给定一个字符流，从流中找到第一个非重复字符。您需要随时在 O(1) 时间内告诉第一个非重复字符。
如果我们遵循这里讨论的第一种方法，那么我们需要存储流，以便我们可以在任何时候再次遍历它以找到第一个非重复字符。如果我们使用同一篇文章中讨论的扩展方法，每次查询第一个非重复元素时，我们都需要遍历计数数组。我们可以随时从流中找到第一个非重复字符，而无需遍历任何数组。
该想法是使用一个DLL（d oubly大号着墨大号IST）以有效地获得从流的第一个非重复的字符。 DLL 按顺序包含所有非重复字符，即DLL 的头部包含第一个非重复字符，第二个节点包含第二个非重复字符，依此类推。
我们还维护了两个数组：一个数组是维护已经被访问过两次或多次的字符，我们称之为repeat[]，另一个数组是一个指向链表节点的指针数组，我们称之为inDLL[]。两个数组的大小等于字母表大小，通常为 256。
- 创建一个空的 DLL。还要创建两个大小为 256 的数组 inDLL[] 和重复的 []。 inDLL 是指向 DLL 节点的指针数组。 repeat[] 是一个布尔数组，如果 x 重复两次或多次，repeat[x] 为真，否则为假。如果 DLL 中存在字符x，则 inDLL[x] 包含指向 DLL 节点的指针，否则为 NULL。
- 将 inDLL[] 的所有条目初始化为 NULL，并将重复 [] 初始化为 false。
- 要获取第一个非重复字符，请返回 DLL 开头的字符。
- 以下是处理流中新字符“x”的步骤。
- 如果repeat[x] 为真，忽略这个字符（x 已经在流中重复了两次或更多次）
- 如果repeat[x] 为false 并且inDLL[x] 为NULL（第一次看到x）。将 x 附加到 DLL 并将新 DLL 节点的地址存储在 inDLL[x] 中。
- 如果repeat[x] 为false 并且inDLL[x] 不是NULL（第二次看到x）。使用 inDLL[x] 获取 x 的 DLL 节点并删除该节点。此外，将 inDLL[x] 标记为 NULL 并将重复 [x] 标记为 true。
请注意，如果我们维护尾指针，将新节点附加到 DLL 是O(1)操作。从 DLL 中删除节点也是O(1) 。因此，添加新字符和查找第一个非重复字符这两个操作都需要O(1)时间。
下图是上述方法的试运行：
下面是上述方法的实现：
C++
// A C++ program to find first
// non-repeating character
// from a stream of characters
#include
#define MAX_CHAR 256
using namespace std;
// A linked list node
struct node {
char a;
struct node *next, *prev;
};
// A utility function to append a character x at the end
// of DLL. Note that the function may change head and tail
// pointers, that is why pointers to these pointers are
// passed.
void appendNode(struct node** head_ref,
struct node** tail_ref, char x)
{
struct node* temp = new node;
temp->a = x;
temp->prev = temp->next = NULL;
if (*head_ref == NULL) {
*head_ref = *tail_ref = temp;
return;
}
(*tail_ref)->next = temp;
temp->prev = *tail_ref;
*tail_ref = temp;
}
// A utility function to remove a node 'temp' fromt DLL.
// Note that the function may change head and tail pointers,
// that is why pointers to these pointers are passed.
void removeNode(struct node** head_ref,
struct node** tail_ref, struct node* temp)
{
if (*head_ref == NULL)
return;
if (*head_ref == temp)
*head_ref = (*head_ref)->next;
if (*tail_ref == temp)
*tail_ref = (*tail_ref)->prev;
if (temp->next != NULL)
temp->next->prev = temp->prev;
if (temp->prev != NULL)
temp->prev->next = temp->next;
delete (temp);
}
void findFirstNonRepeating()
{
// inDLL[x] contains pointer to
// a DLL node if x is present
// in DLL. If x is not present, then inDLL[x] is NULL
struct node* inDLL[MAX_CHAR];
// repeated[x] is true if x is repeated two or more
// times. If x is not seen so far or x is seen only
// once. then repeated[x] is false
bool repeated[MAX_CHAR];
// Initialize the above two arrays
struct node *head = NULL, *tail = NULL;
for (int i = 0; i < MAX_CHAR; i++) {
inDLL[i] = NULL;
repeated[i] = false;
}
// Let us consider following stream and see the process
char stream[] = "geeksforgeeksandgeeksquizfor";
for (int i = 0; stream[i]; i++) {
char x = stream[i];
cout << "Reading " << x << " from stream \n";
// We process this character only if it has not
// occurred or occurred only once. repeated[x] is
// true if x is repeated twice or more.s
if (!repeated[x]) {
// If the character is not in DLL, then add this
// at the end of DLL.
if (inDLL[x] == NULL) {
appendNode(&head, &tail, stream[i]);
inDLL[x] = tail;
}
else // Otherwise remove this character from DLL
{
removeNode(&head, &tail, inDLL[x]);
inDLL[x] = NULL;
repeated[x]
= true; // Also mark it as repeated
}
}
// Print the current first non-repeating character
// from stream
if (head != NULL)
cout << "First non-repeating character so far "
"is "
<< head->a << endl;
}
}
/* Driver code */
int main()
{
findFirstNonRepeating();
return 0;
} Java
// A Java program to find first non-repeating character
// from a stream of characters
import java.util.ArrayList;
import java.util.List;
public class NonReapeatingC {
final static int MAX_CHAR = 256;
static void findFirstNonRepeating()
{
// inDLL[x] contains pointer to a DLL node if x is
// present in DLL. If x is not present, then
// inDLL[x] is NULL
List inDLL = new ArrayList();
// repeated[x] is true if x is repeated two or more
// times. If x is not seen so far or x is seen only
// once. then repeated[x] is false
boolean[] repeated = new boolean[MAX_CHAR];
// Let us consider following stream and see the
// process
String stream = "geeksforgeeksandgeeksquizfor";
for (int i = 0; i < stream.length(); i++) {
char x = stream.charAt(i);
System.out.println("Reading " + x
+ " from stream \n");
// We process this character only if it has not
// occurred or occurred only once. repeated[x]
// is true if x is repeated twice or more.s
if (!repeated[x]) {
// If the character is not in DLL, then add
// this at the end of DLL.
if (!(inDLL.contains(x))) {
inDLL.add(x);
}
else // Otherwise remove this character from
// DLL
{
inDLL.remove((Character)x);
repeated[x]
= true; // Also mark it as repeated
}
}
// Print the current first non-repeating
// character from stream
if (inDLL.size() != 0) {
System.out.print(
"First non-repeating character so far is ");
System.out.println(inDLL.get(0));
}
}
}
/* Driver code */
public static void main(String[] args)
{
findFirstNonRepeating();
}
}
// This code is contributed by Sumit Ghosh Python3
# A Python program to find first non-repeating character from
# a stream of characters
MAX_CHAR = 256
def findFirstNonRepeating():
# inDLL[x] contains pointer to a DLL node if x is present
# in DLL. If x is not present, then inDLL[x] is NULL
inDLL = [] * MAX_CHAR
# repeated[x] is true if x is repeated two or more times.
# If x is not seen so far or x is seen only once. then
# repeated[x] is false
repeated = [False] * MAX_CHAR
# Let us consider following stream and see the process
stream = "geekforgeekandgeeksandquizfor"
for i in range(len(stream)):
x = stream[i]
print ("Reading " + x + " from stream")
# We process this character only if it has not occurred
# or occurred only once. repeated[x] is true if x is
# repeated twice or more.s
if not repeated[ord(x)]:
# If the character is not in DLL, then add this
# at the end of DLL
if not x in inDLL:
inDLL.append(x)
else:
inDLL.remove(x)
repeated[ord(x)] = True
if len(inDLL) != 0:
print ("First non-repeating character so far is ")
print (str(inDLL[0]))
# Driver program
findFirstNonRepeating()
# This code is contributed by BHAVYA JAINC#
// A C# program to find first non-repeating character
// from a stream of characters
using System;
using System.Collections.Generic;
public class NonReapeatingC {
readonly static int MAX_CHAR = 256;
static void findFirstNonRepeating()
{
// inDLL[x] contains pointer to a DLL node if x is present
// in DLL. If x is not present, then inDLL[x] is NULL
List inDLL = new List();
// repeated[x] is true if x is repeated two or more times.
// If x is not seen so far or x is seen only once. then
// repeated[x] is false
bool[] repeated = new bool[MAX_CHAR];
// Let us consider following stream and see the process
String stream = "geeksforgeeksandgeeksquizfor";
for (int i = 0; i < stream.Length; i++) {
char x = stream[i];
Console.WriteLine("Reading " + x + " from stream \n");
// We process this character only if it has not occurred
// or occurred only once. repeated[x] is true if x is
// repeated twice or more.s
if (!repeated[x]) {
// If the character is not in DLL, then add this at
// the end of DLL.
if (!(inDLL.Contains(x))) {
inDLL.Add(x);
}
else // Otherwise remove this character from DLL
{
inDLL.Remove((char)x);
repeated[x] = true; // Also mark it as repeated
}
}
// Print the current first non-repeating character from
// stream
if (inDLL.Count != 0) {
Console.Write("First non-repeating character so far is ");
Console.WriteLine(inDLL[0]);
}
}
}
/* Driver code*/
public static void Main(String[] args)
{
findFirstNonRepeating();
}
}
// This code has been contributed by 29AjayKumar Javascript
输出：
Reading g from stream
First non-repeating character so far is g
Reading e from stream
First non-repeating character so far is g
Reading e from stream
First non-repeating character so far is g
Reading k from stream
First non-repeating character so far is g
Reading s from stream
First non-repeating character so far is g
Reading f from stream
First non-repeating character so far is g
Reading o from stream
First non-repeating character so far is g
Reading r from stream
First non-repeating character so far is g
Reading g from stream
First non-repeating character so far is k
Reading e from stream
First non-repeating character so far is k
Reading e from stream
First non-repeating character so far is k
Reading k from stream
First non-repeating character so far is s
Reading s from stream
First non-repeating character so far is f
Reading a from stream
First non-repeating character so far is f
Reading n from stream
First non-repeating character so far is f
Reading d from stream
First non-repeating character so far is f
Reading g from stream
First non-repeating character so far is f
Reading e from stream
First non-repeating character so far is f
Reading e from stream
First non-repeating character so far is f
Reading k from stream
First non-repeating character so far is f
Reading s from stream
First non-repeating character so far is f
Reading q from stream
First non-repeating character so far is f
Reading u from stream
First non-repeating character so far is f
Reading i from stream
First non-repeating character so far is f
Reading z from stream
First non-repeating character so far is f
Reading f from stream
First non-repeating character so far is o
Reading o from stream
First non-repeating character so far is r
Reading r from stream
First non-repeating character so far is a如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。