Given a string, find the length of the longest substring without repeating characters.
Example 1:
Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:
Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:
Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
给定一个字符串，求出它最长的没有重复字符的子串的长度。
法一：定义字符串str、rst，遍历给定的字符串s，每遍历一个字符c，判断它是否在字符串str中存在，若不存在，直接添加到字符串str末尾，若存在，比较字符串str与rst的长度，若前者比后者长，将字符串赋给rst，并抹去字符串str从0到字符c处的所有字符。当遍历结束时，字符串rst即为给定字符串s的最长不含重复字符的子串。当当前最长子串（即字符串rst）的长度比 剩余字符数+当前字符串长度（即字符串str） 还长时，没必要继续遍历下去，因为已经不可能存在比字符串rst还长的子串了。代码如下：
int lengthOfLongestSubstring(std::string s) {
size_t pos = 0;
std::string str, rst;
int cur = 0, total = s.length();
for (auto c : s)
{
pos = str.find(c);
if (pos != std::string::npos)
{
if (str.length() > rst.length())
rst = str;
str = str.substr(pos + 1);
if (rst.length() >= str.length() + total - cur)
break;
}
str.append(1, c);
cur++;
}
int cur_len = str.length();
int record_len = rst.length();
return cur_len > record_len ? cur_len : record_len;
}
法二：法一中遍历给定字符串是无法避免的，查找字符str是否含有字符c可以用HashMap来做，因为最终只需要最长不含重复字符的子串的长度，我们甚至不需要保存临时字符串str，改进后的代码如下：
int lengthOfLongestSubstring(std::string s) {
std::unordered_set<char> set;
int left = 0,length = 0, cnt = 0, cur = 0, total = s.length();
for (char c : s)
{
if (set.find(c) != set.end())
{
cnt = set.size();
if (cnt > length)
length = cnt;
char ch = ‘\0‘;
while (set.find(c) != set.end())
{
ch = s.at(left);
auto iter = set.find(ch);
if (iter != set.end())
set.erase(iter);
left++;
}
}
set.insert(c);
cur++;
if (length >= (int)set.size() + total - cur)
break;
}
cnt = set.size();
return length > cnt ? length : cnt;
}
3. Longest Substring Without Repeating Characters
原文：https://www.cnblogs.com/big-potato/p/12263000.html