【leetcode】3. Longest Substring Without Repeating Characters (medium)

最长无重复子串算法

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本文探讨了如何在给定字符串中找到最长的无重复字符子串，通过两个示例详细解析了算法实现过程，包括使用哈希表进行字符位置记录，以及在遇到重复字符时如何调整搜索范围。

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.

// 求最大无重复字符的子串
注意，是无重复字符的，最大长度子串长度。

//3. Longest Substring Without Repeating Characters
// 求最大无重复字符的子串
int Solution::lengthOfLongestSubstring(string s)
{
	unordered_map<char, int> hash;
	int first = 0;
	int res = 0;
	for (int i = 0; i < s.size(); ++i)
	{
		char cur = s[i];
		if (hash.size() == 0 || hash.find(cur) == hash.end())
		{
			hash.insert({ cur, i });
			res = hash.size() > res ? hash.size() : res;
		}
		else
		{
			int index = hash.find(cur)->second;
			auto it = hash.begin();
			while (it != hash.end())
			{
				if (it->second <= index) // 删掉前面的
				{
					it = hash.erase(it);
					hash.insert({ cur, i });
				}
				else
					++it;
			}
		}
	}
	return res;
}

效率一般，不过是ac了。

Runtime: 68 ms, faster than 22.72% of C++ online submissions for Longest Substring Without Repeating Characters.
Memory Usage: 20.5 MB, less than 20.56% of C++ online submissions for Longest Substring Without Repeating Characters.

2020.6.16

int lengthOfLongestSubstring(string s) {
	int res = 0;
	unordered_map<char, int> hash;

	for (int i = 0; i < s.size(); i++)
	{
		if (!hash.empty() && hash.find(s[i]) != hash.cend())
		{
			int index = hash.find(s[i])->second;
			auto it = hash.begin();
			while (it != hash.end())
			{
				if (it->second <= index)
				{
					it = hash.erase(it);
				}
				else
				{
					it++;
				}
			}
			hash.insert({ s[i], i });
		}
		else{
			hash.insert(std::make_pair(s[i], i));
			res = res > hash.size() ? res : hash.size();
		}
	}
	return res;
}