LeetCode 3---Longest Substring Without Repeating Characters-优快云博客

本文链接：https://blog.youkuaiyun.com/liuwg1226/article/details/126684149

本文探讨了求解字符串中最长无重复字符子串的两种常见算法——双指针法。首先介绍了Solution1中基于StringBuilder的实现，然后展示了Solution2中利用HashSet的数据结构优化。通过实例解析，帮助读者理解如何在字符串处理问题中使用这些技巧。

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Given a string s, find the length of the longest substring without repeating characters.

Example 1:

Input: s = “abcabcbb”
Output: 3
Explanation: The answer is “abc”, with the length of 3.

Example 2:

Input: s = “bbbbb”
Output: 1
Explanation: The answer is “b”, with the length of 1。

Example 3:

Input: s = “pwwkew”
Output: 3
Explanation: The answer is “wke”, with the length of 3.
Notice that the answer must be a substring, “pwke” is a subsequence and not a substring.

Constraints:

0 <= s.length <= 5 * 104
s consists of English letters, digits, symbols and spaces.

Solution1 dual circulation

class Solution {
    public int lengthOfLongestSubstring(String s) {      
      
        if (s.length() == 0) {
            return 0;
        }

        int currentLongest = 1;
        
        for (int i = 0; i < s.length(); i++) {
            StringBuilder sb = new StringBuilder();
            sb.append(s.charAt(i));
            for (int j = i + 1; j < s.length(); j++) {                
                if (sb.toString().contains(Character.toString(s.charAt(j)))) {
                    break;
                } else {
                    sb.append(s.charAt(j));
                }              
            }
            currentLongest = Math.max(currentLongest, sb.length());
        }
        return currentLongest;
    }
}

Solution2 dual circulation

class Solution {
    public int lengthOfLongestSubstring(String s) {      
      
        Set<Character> set = new HashSet<Character>();
		int left = 0 ,right = 0, res = 0;
		while(right < s.length()){
			char c = s.charAt(right++);
			while(set.contains(c)){
				set.remove(s.charAt(left++));
			}
			set.add(c);
			res = Math.max(res, right - left);
		}
		return res;

    }
}