打印无重复字符的最长子字符串
原文：https://www.geeksforgeeks.org/print-longest-substring-without-repeating-characters/
给定一个字符串，打印最长的子字符串，而不重复字符。 例如，ABDEFGABEF的没有重复字符的最长子串是BDEFGA和DEFGAB，长度为 6。对于BBBB，最长子串是B，长度为 1。所需的时间复杂度为O(n)，其中n是字符串的长度。
先决条件：最长子串的长度，不包含重复字符
示例：
Input : GEEKSFORGEEKS
Output : EKSFORG
Input : ABDEFGABEF
Output : BDEFGA
Recommended : Please try your approach first on IDE and then look at the solution.
方法：想法是遍历字符串，对于每个已访问的字符，将其最后一次出现存储在哈希表中（此处unordered_map用作哈希，键作为字符，值作为其最后位置）。 变量st存储当前子串的起点，maxlen存储最大长度子串的长度，start存储最大长度子串的起始索引。 在遍历字符串时，请检查哈希表中是否存在当前字符。 如果不存在，则将当前字符存储在哈希表中，并将值作为当前索引。 如果哈希表中已经存在该字符，则意味着当前字符可以在当前子字符串中重复。 对于此检查，字符的上一次出现是在当前子字符串的起点st之前还是之后。 如果它在st之前，则仅更新哈希表中的值。 如果在st之后，则找到当前子串currlen的长度为i - st，其中i是当前索引。 比较currlen和maxlen。 如果maxlen小于currlen，则将maxlen更新为currlen并从st开始。 完整遍历字符串后，所需的最长子字符串（不重复字符）为s[start]到s [start + maxlen - 1]。
实现：
C++
// C++ program to find and print longest
// substring without repeating characters.
#include <bits/stdc++.h>
using namespace std;
// Function to find and print longest
// substring without repeating characters.
string findLongestSubstring(string str)
{
int i;
int n = str.length();
// starting point of current substring.
int st = 0;
// length of current substring.
int currlen;
// maximum length substring without repeating
// characters.
int maxlen = 0;
// starting index of maximum length substring.
int start;
// Hash Map to store last occurrence of each
// already visited character.
unordered_map<char, int> pos;
// Last occurrence of first character is index 0;
pos[str[0]] = 0;
for (i = 1; i < n; i++) {
// If this character is not present in hash,
// then this is first occurrence of this
// character, store this in hash.
if (pos.find(str[i]) == pos.end())
pos[str[i]] = i;
else {
// If this character is present in hash then
// this character has previous occurrence,
// check if that occurrence is before or after
// starting point of current substring.
if (pos[str[i]] >= st) {
// find length of current substring and
// update maxlen and start accordingly.
currlen = i - st;
if (maxlen < currlen) {
maxlen = currlen;
start = st;
}
// Next substring will start after the last
// occurrence of current character to avoid
// its repetition.
st = pos[str[i]] + 1;
}
// Update last occurrence of
// current character.
pos[str[i]] = i;
}
}
// Compare length of last substring with maxlen and
// update maxlen and start accordingly.
if (maxlen < i - st) {
maxlen = i - st;
start = st;
}
// The required longest substring without
// repeating characters is from str[start]
// to str[start+maxlen-1].
return str.substr(start, maxlen);
}
// Driver function
int main()
{
string str = "GEEKSFORGEEKS";
cout << findLongestSubstring(str);
return 0;
}
Java
// Java program to find
// and print longest substring
// without repeating characters.
import java.util.*;
class GFG{
// Function to find and print longest
// substring without repeating characters.
public static String findLongestSubstring(String str)
{
int i;
int n = str.length();
// Starting point
// of current substring.
int st = 0;
// length of
// current substring.
int currlen = 0;
// maximum length
// substring without
// repeating characters.
int maxlen = 0;
// starting index of
// maximum length substring.
int start = 0;
// Hash Map to store last
// occurrence of each
// already visited character.
HashMap<Character,
Integer> pos = new HashMap<Character,
Integer>();
// Last occurrence of first
// character is index 0;
pos.put(str.charAt(0), 0);
for (i = 1; i < n; i++)
{
// If this character is not present in hash,
// then this is first occurrence of this
// character, store this in hash.
if (!pos.containsKey(str.charAt(i)))
{
pos.put(str.charAt(i), i);
}
else
{
// If this character is present
// in hash then this character
// has previous occurrence,
// check if that occurrence
// is before or after starting
// point of current substring.
if (pos.get(str.charAt(i)) >= st)
{
// find length of current
// substring and update maxlen
// and start accordingly.
currlen = i - st;
if (maxlen < currlen)
{
maxlen = currlen;
start = st;
}
// Next substring will start
// after the last occurrence
// of current character to avoid
// its repetition.
st = pos.get(str.charAt(i)) + 1;
}
// Update last occurrence of
// current character.
pos.replace(str.charAt(i), i);
}
}
// Compare length of last
// substring with maxlen and
// update maxlen and start
// accordingly.
if (maxlen < i - st)
{
maxlen = i - st;
start = st;
}
// The required longest
// substring without
// repeating characters
// is from str[start]
// to str[start+maxlen-1].
return str.substring(start,
start +
maxlen);
}
// Driver Code
public static void main(String[] args)
{
String str = "GEEKSFORGEEKS";
System.out.print(findLongestSubstring(str));
}
}
// This code is contributed by divyeshrabadiya07
Python3
# Python3 program to find and print longest
# substring without repeating characters.
# Function to find and print longest
# substring without repeating characters.
def findLongestSubstring(string):
n = len(string)
# starting point of current substring.
st = 0
# maximum length substring without
# repeating characters.
maxlen = 0
# starting index of maximum
# length substring.
start = 0
# Hash Map to store last occurrence
# of each already visited character.
pos = {}
# Last occurrence of first
# character is index 0
pos[string[0]] = 0
for i in range(1, n):
# If this character is not present in hash,
# then this is first occurrence of this
# character, store this in hash.
if string[i] not in pos:
pos[string[i]] = i
else:
# If this character is present in hash then
# this character has previous occurrence,
# check if that occurrence is before or after
# starting point of current substring.
if pos[string[i]] >= st:
# find length of current substring and
# update maxlen and start accordingly.
currlen = i - st
if maxlen < currlen:
maxlen = currlen
start = st
# Next substring will start after the last
# occurrence of current character to avoid
# its repetition.
st = pos[string[i]] + 1
# Update last occurrence of
# current character.
pos[string[i]] = i
# Compare length of last substring with maxlen
# and update maxlen and start accordingly.
if maxlen < i - st:
maxlen = i - st
start = st
# The required longest substring without
# repeating characters is from string[start]
# to string[start+maxlen-1].
return string[start : start + maxlen]
# Driver Code
if __name__ == "__main__":
string = "GEEKSFORGEEKS"
print(findLongestSubstring(string))
# This code is contributed by Rituraj Jain
C
// C# program to find
// and print longest substring
// without repeating characters.
using System;
using System.Collections.Generic;
class GFG{
// Function to find and
// print longest substring
// without repeating characters.
public static String findlongestSubstring(String str)
{
int i;
int n = str.Length;
// Starting point
// of current substring.
int st = 0;
// length of
// current substring.
int currlen = 0;
// maximum length
// substring without
// repeating characters.
int maxlen = 0;
// starting index of
// maximum length substring.
int start = 0;
// Hash Map to store last
// occurrence of each
// already visited character.
Dictionary<char,
int> pos = new Dictionary<char,
int>();
// Last occurrence of first
// character is index 0;
pos.Add(str[0], 0);
for (i = 1; i < n; i++)
{
// If this character is not present in hash,
// then this is first occurrence of this
// character, store this in hash.
if (!pos.ContainsKey(str[i]))
{
pos.Add(str[i], i);
}
else
{
// If this character is present
// in hash then this character
// has previous occurrence,
// check if that occurrence
// is before or after starting
// point of current substring.
if (pos[str[i]] >= st)
{
// find length of current
// substring and update maxlen
// and start accordingly.
currlen = i - st;
if (maxlen < currlen)
{
maxlen = currlen;
start = st;
}
// Next substring will start
// after the last occurrence
// of current character to avoid
// its repetition.
st = pos[str[i]] + 1;
}
// Update last occurrence of
// current character.
pos[str[i]] = i;
}
}
// Compare length of last
// substring with maxlen and
// update maxlen and start
// accordingly.
if (maxlen < i - st)
{
maxlen = i - st;
start = st;
}
// The required longest
// substring without
// repeating characters
// is from str[start]
// to str[start+maxlen-1].
return str.Substring(start,
maxlen);
}
// Driver Code
public static void Main(String[] args)
{
String str = "GEEKSFORGEEKS";
Console.Write(findlongestSubstring(str));
}
}
// This code is contributed by shikhasingrajput
输出：
EKSFORG
时间复杂度：O(n)。
辅助空间：O(n)。
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