[3, Medium, C++] Longest Substring Without Repeating Characters

本文介绍了一种求解字符串中最长无重复字符子串长度的高效算法。通过使用字符特征数组,该算法能快速定位重复字符并更新子串长度,实现O(n)的时间复杂度。

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Problem:

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.

Analysis:

To solve this question, the program first caches a set of distinct characters of the string and runs from the first one to the last one to check whether a repeated character of the set is met. If such character is met, remove all ones before the repeated character. The key point of this method is to locate the repeated one in the quickest way.The method here is a typical one. We use a characteristic array whose index is the ascii value of a character and value is the index this character. For each character of the string, if the corresponding characteristic value of this character is greater or equal of the current sub-string, it means that a repeated character is found. Then, update the local_max_length of the sub-string and new start-checking-point. 

Sample Code:

    int lengthOfLongestSubstring(string s) {
        if(s.size() <= 1)
            return s.size();
            
        int max_length = 0;
        vector<int> char_seq_index(256, -1);
        int start_point = 0;
        int local_max_length = 0;
        for(int i = 0; i < s.size(); ++i) {
            if(char_seq_index[s[i]] >= start_point) {
                local_max_length -= char_seq_index[s[i]] + 1 - start_point;
                start_point = char_seq_index[s[i]] + 1;
            }
            
            char_seq_index[s[i]] = i;
            ++local_max_length;
            
            if(local_max_length > max_length)
                max_length = local_max_length;
        }
        
        return max_length;
    }


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