Leetcode 3 Longest Substring without repeating characters

最新推荐文章于 2025-05-18 01:04:12 发布

jenny1000000

最新推荐文章于 2025-05-18 01:04:12 发布

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本文探讨了如何寻找字符串中最长的无重复字符子串，提供了两种解决方案：暴力破解法和滑动窗口法。通过实例解析算法思路，帮助读者理解并掌握最长无重复子串的求解方法。

Longest Substring without repeating characters

Question

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.

Naïve solution: brute force

The idea is to check every substring of the string, and use a set to check if the substring contains duplicate.

class Solution
{
public:
	int lengthOfLongestSubstring(string s)
	{
		int i, j, k;
		int longest = 0;
        for (i = 0; i<s.length() - 1; i++)
            for (j = i + 1; j<s.length(); j++)
            {
                set<char> substring;
                for (k = i; k <= j; k++)
                {
                    if (substring.find(s[k]) != substring.end())
                    {
                        longest = max(longest, int(substring.size()));
                        break;
                    }
                    else
                        substring.insert(s[k]);
                }
                //这个小心！ e.g. "pwwkew"
                longest = max(longest, int(substring.size()));
            }

		return longest;
	}
};

Sliding Window

在这里插入图片描述

A window is represented by $[i, j)$ , containing a substring.

The idea is to use a sliding window to store the characters in current window, if s[j] is not in the set, then slide j further. Otherwise we find the maximum size of the current substring starting from i. Repeat for a new i.

class Solution 
{
public:
    int lengthOfLongestSubstring(string s) 
    {
        int i=0, j=0;
        int longest = 0;
        unordered_map <char, int> characters;

        while( i<s.length() && j<s.length() )
        {
            if(characters.find(s[j])==characters.end() || characters[s[j]]<i || characters[s[j]]>=j)
            {
                //此时map种没有这个值或者位置在window之外
                characters[s[j]] = j;
                j++;
            }
            else
            {
                if((j-i)>longest)
                    longest = j-i;
                i = characters[s[j]]+1;
                j = i;
            }
        }

        if((j-i)>longest)
            longest = j-i;

        return longest;
    }
};