使用一次字符串遍历的第一个非重复字符|设置 2
给定一个字符串，找出其中第一个不重复的字符。例如，如果输入字符串是“GeeksforGeeks”，那么输出应该是“f”，如果输入字符串是“GeeksQuiz”，那么输出应该是“G”。
我们在给定一个字符串中讨论了两种解决方案，找到它的第一个非重复字符。在这篇文章中，讨论了进一步优化的解决方案（在上一篇文章的方法 2 之上）。这个想法是优化空间。我们不是使用对来存储计数和索引，而是使用单个元素来存储索引，如果一个元素出现一次，则存储一个负值。
C++
// CPP program to find first non-repeating
// character using 1D array and one traversal.
#include
using namespace std;
#define NO_OF_CHARS 256
/* The function returns index of the first
non-repeating character in a string. If
all characters are repeating then
returns INT_MAX */
int firstNonRepeating(char* str)
{
// Initialize all characters as
// absent.
int arr[NO_OF_CHARS];
for (int i = 0; i < NO_OF_CHARS; i++)
arr[i] = -1;
// After below loop, the value of
// arr[x] is going to be index of
// of x if x appears only once. Else
// the value is going to be either
// -1 or -2.
for (int i = 0; str[i]; i++) {
if (arr[str[i]] == -1)
arr[str[i]] = i;
else
arr[str[i]] = -2;
}
int res = INT_MAX;
for (int i = 0; i < NO_OF_CHARS; i++)
// If this character occurs only
// once and appears before the
// current result, then update the
// result
if (arr[i] >= 0)
res = min(res, arr[i]);
return res;
}
/* Driver program to test above function */
int main()
{
char str[] = "geeksforgeeks";
int index = firstNonRepeating(str);
if (index == INT_MAX)
cout <<"Either all characters are "
"repeating or string is empty";
else
cout <<"First non-repeating character"
" is "<< str[index];
return 0;
}
// This code is contributed by shivanisinghss210 C
// CPP program to find first non-repeating
// character using 1D array and one traversal.
#include
#include
#include
#define NO_OF_CHARS 256
/* The function returns index of the first
non-repeating character in a string. If
all characters are repeating then
returns INT_MAX */
int firstNonRepeating(char* str)
{
// Initialize all characters as
// absent.
int arr[NO_OF_CHARS];
for (int i = 0; i < NO_OF_CHARS; i++)
arr[i] = -1;
// After below loop, the value of
// arr[x] is going to be index of
// of x if x appears only once. Else
// the value is going to be either
// -1 or -2.
for (int i = 0; str[i]; i++) {
if (arr[str[i]] == -1)
arr[str[i]] = i;
else
arr[str[i]] = -2;
}
int res = INT_MAX;
for (int i = 0; i < NO_OF_CHARS; i++)
// If this character occurs only
// once and appears before the
// current result, then update the
// result
if (arr[i] >= 0)
res = min(res, arr[i]);
return res;
}
/* Driver program to test above function */
int main()
{
char str[] = "geeksforgeeks";
int index = firstNonRepeating(str);
if (index == INT_MAX)
printf("Either all characters are "
"repeating or string is empty");
else
printf("First non-repeating character"
" is %c", str[index]);
return 0;
} Java
// Java program to find first
// non-repeating character
// using 1D array and one
// traversal.
import java.io.*;
import java.util.*;
import java.lang.*;
class GFG
{
/* The function returns index
of the first non-repeating
character in a string. If
all characters are repeating
then returns INT_MAX */
static int firstNonRepeating(String str)
{
int NO_OF_CHARS = 256;
// Initialize all characters
// as absent.
int arr[] = new int[NO_OF_CHARS];
for (int i = 0;
i < NO_OF_CHARS; i++)
arr[i] = -1;
// After below loop, the
// value of arr[x] is going
// to be index of x if x
// appears only once. Else
// the value is going to be
// either -1 or -2.
for (int i = 0;
i < str.length(); i++)
{
if (arr[str.charAt(i)] == -1)
arr[str.charAt(i)] = i;
else
arr[str.charAt(i)] = -2;
}
int res = Integer.MAX_VALUE;
for (int i = 0; i < NO_OF_CHARS; i++)
// If this character occurs
// only once and appears before
// the current result, then
// update the result
if (arr[i] >= 0)
res = Math.min(res, arr[i]);
return res;
}
// Driver Code
public static void main(String args[])
{
String str = "geeksforgeeks";
int index = firstNonRepeating(str);
if (index == Integer.MAX_VALUE)
System.out.print("Either all characters are " +
"repeating or string is empty");
else
System.out.print("First non-repeating character"+
" is " + str.charAt(index));
}
}Python3
'''
Python3 implementation to find non repeating character using
1D array and one traversal'''
import math as mt
NO_OF_CHARS = 256
'''
The function returns index of the first
non-repeating character in a string. If
all characters are repeating then
returns INT_MAX '''
def firstNonRepeating(string):
#initialize all character as absent
arr=[-1 for i in range(NO_OF_CHARS)]
'''
After below loop, the value of
arr[x] is going to be index of
of x if x appears only once. Else
the value is going to be either
-1 or -2.'''
for i in range(len(string)):
if arr[ord(string[i])]==-1:
arr[ord(string[i])]=i
else:
arr[ord(string[i])]=-2
res=10**18
for i in range(NO_OF_CHARS):
'''
If this character occurs only
once and appears before the
current result, then update the
result'''
if arr[i]>=0:
res=min(res,arr[i])
return res
#Driver prohram to test above function
string="geeksforgeeks"
index=firstNonRepeating(string)
if index==10**18:
print("Either all characters are repeating or string is empty")
else:
print("First non-repeating character is",string[index])
#this code is contributed by Mohit Kumar 29C#
// C# program to find first
// non-repeating character
// using 1D array and one
// traversal.
using System;
class GFG
{
/* The function returns index
of the first non-repeating
character in a string. If
all characters are repeating
then returns INT_MAX */
static int firstNonRepeating(String str)
{
int NO_OF_CHARS = 256;
// Initialize all characters
// as absent.
int []arr = new int[NO_OF_CHARS];
for (int i = 0; i < NO_OF_CHARS; i++)
arr[i] = -1;
// After below loop, the
// value of arr[x] is going
// to be index of x if x
// appears only once. Else
// the value is going to be
// either -1 or -2.
for (int i = 0; i < str.Length; i++)
{
if (arr[str[i]] == -1)
arr[str[i]] = i;
else
arr[str[i]] = -2;
}
int res = int.MaxValue;
for (int i = 0; i < NO_OF_CHARS; i++)
// If this character occurs
// only once and appears before
// the current result, then
// update the result
if (arr[i] >= 0)
res = Math.Min(res, arr[i]);
return res;
}
// Driver Code
public static void Main()
{
String str = "geeksforgeeks";
int index = firstNonRepeating(str);
if (index == int.MaxValue)
Console.Write("Either all characters are " +
"repeating or string is empty");
else
Console.Write("First non-repeating character"+
" is " + str[index]);
}
}
// This code is contributed by Rajput-JiJavascript
Python3
# Python implementation of
# above approach
from collections import Counter
def makeHashMap(string):
# Use Counter to make a frequency
# dictionary of characters in a string.
d1 = Counter(string)
# Update value of each key such that
# if frequency of frequency of a key
# is equal to 1, then it is set to 0,
# else set the value equal to None
d1 = {(key): (0 if d1[key] == 1 else None)
for key, value in d1.items()}
return d1
def firstNotRepeatingCharacter(s):
d = makeHashMap(s)
# variable to store the first
# non repeating character.
nonRep = None
# Traversing through the string only once.
for i in s:
if d[i] is not None:
'''
As soon as the first character in the string is
found whose value in the HashMap is "not None",
that is our first non-repeating character.
So we store it in nonRep and break out of the
loop. Thus saving on time.
'''
nonRep = i
break
if nonRep is not None:
return nonRep
else:
# If there are no non-repeating characters
# then "_" is returned
return str("_")
# Driver Code
print(firstNotRepeatingCharacter('bbcdcca'))
# This code is contributed by Vivek Nigam (@viveknigam300)Java
// Java program for the above approach
public class first_non_repeating {
static int firstNonRepeating(String str)
{
boolean flag = false;
int index = -1;
for (int i = 0; i < s.length(); i++) {
// Here I am replacing a character with "" and
// then finding the length after replacing and
// then subtracting length of that replaced
// string from the total length of the original
// string
int count_occurence
= s.length()
- s.replace(
Character.toString(s.charAt(i)),
"")
.length();
if (count_occurence == 1) {
flag = true;
index = i;
break;
}
}
if (flag)
System.out.println(
"First non repeating character is "
+ s.charAt(index));
else
System.out.println(
"There is no non-repeating character
is present in the string");
}
// Driver Code
public static void main(String arg[])
{
String s = "GeeksforGeeks";
firstNonRepeating(s);
}
}
// This Solution is contributed by Sourabh Sharma.输出
First non-repeating character is f时间复杂度： O(n)
替代实施
这是使用 HashMap 或散列技术编码的。
如果元素（或键）在字符串中重复，则 HashMap（或字典）将该键的值更改为无。
这样，我们稍后将只找到值为“非无”的键。
Python3
# Python implementation of
# above approach
from collections import Counter
def makeHashMap(string):
# Use Counter to make a frequency
# dictionary of characters in a string.
d1 = Counter(string)
# Update value of each key such that
# if frequency of frequency of a key
# is equal to 1, then it is set to 0,
# else set the value equal to None
d1 = {(key): (0 if d1[key] == 1 else None)
for key, value in d1.items()}
return d1
def firstNotRepeatingCharacter(s):
d = makeHashMap(s)
# variable to store the first
# non repeating character.
nonRep = None
# Traversing through the string only once.
for i in s:
if d[i] is not None:
'''
As soon as the first character in the string is
found whose value in the HashMap is "not None",
that is our first non-repeating character.
So we store it in nonRep and break out of the
loop. Thus saving on time.
'''
nonRep = i
break
if nonRep is not None:
return nonRep
else:
# If there are no non-repeating characters
# then "_" is returned
return str("_")
# Driver Code
print(firstNotRepeatingCharacter('bbcdcca'))
# This code is contributed by Vivek Nigam (@viveknigam300)
输出
d时间复杂度： O(N)
方法3：下面提到了另一种不使用任何hashmap或数组的简单方法。我们只需使用单个 for 循环就可以找到第一个不重复的字符。
另一种方法：
要计算字符的频率，我们可以执行以下步骤：
一个字符的频率 = length_of_string – length_of_string_without_that _character
例如：给字符串是“helloh” ，我们想计算字符“h”的频率，所以使用上面的公式我们可以说
字符“h”的频率= 6（字符串长度）- 4（不带“h”的字符串长度）
= 2
因此，通过这种方式，我们可以计算字符串中每个字符的出现频率，然后如果我们发现 count == 1，则表示该字符是字符串中的第一个非重复字符。
实现：上述方法在Java中的实现如下图所示：
Java
// Java program for the above approach
public class first_non_repeating {
static int firstNonRepeating(String str)
{
boolean flag = false;
int index = -1;
for (int i = 0; i < s.length(); i++) {
// Here I am replacing a character with "" and
// then finding the length after replacing and
// then subtracting length of that replaced
// string from the total length of the original
// string
int count_occurence
= s.length()
- s.replace(
Character.toString(s.charAt(i)),
"")
.length();
if (count_occurence == 1) {
flag = true;
index = i;
break;
}
}
if (flag)
System.out.println(
"First non repeating character is "
+ s.charAt(index));
else
System.out.println(
"There is no non-repeating character
is present in the string");
}
// Driver Code
public static void main(String arg[])
{
String s = "GeeksforGeeks";
firstNonRepeating(s);
}
}
// This Solution is contributed by Sourabh Sharma.
输出
First non repeating character is f时间复杂度： O(N)
辅助空间： O(1)