https://leetcode.com/problems/longest-substring-without-repeating-characters-优快云博客

本文探讨了三种不同的算法来寻找字符串中最长的无重复字符子串长度，包括原始的遍历检查法、使用HashSet优化的窗口机制以及利用HashMap实现的高效解决方案。

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我的代码一开始没有考虑字符串为空的特殊情况

public class Test {


    public static int lengthOfLongestSubstring(String s) {
        int longest=0;   //latest char index+1
            byte[] b = s.getBytes();
            int l = b.length;
            byte[] temp = new byte[l];
            for (int i = 0; i < l; i++) {
                temp[0]= b[i]; //start char
                int end=1;
                for (int j=i+1;j<l;j++) {
                    //check b[j] is exist in temp[]
                    boolean isExist=false;
                    for (int k=0;k<end;k++) {
                        if (temp[k] == b[j]) {
                            isExist=true;
                        }
                    }
                    if (isExist == false) {
                        temp[end] = b[j];
                        end++;
                    } else {
                        break;
                    }
                }
                longest=(longest>end)?longest:end;
                if (longest>l-i-1) {
                    break;
                }
            }
            return longest;
    }

    public static void main(String[] args) {
        String s1 = "bbbbb";
        String s2 = "abcabcbb";
        String s3 = "pwwkew";
        int r1 = lengthOfLongestSubstring(s1);
        int r2 = lengthOfLongestSubstring(s2);
        int r3 = lengthOfLongestSubstring(s3);
        System.out.println("r1 = " + r1);
        System.out.println("r2 = " + r2);
        System.out.println("r3 = " + r3);
    }
}

官方答案2:窗口机制，运用了hashSet，速度加快；

  public static int lengthOfLongestSubstring2(String s) {
        int l = s.length();
        HashSet<Character> set = new HashSet<>();
        int ans = 0, i = 0, j = 0;
        while (i < l && j < l) {
            if (!set.contains(s.charAt(j))) {
                set.add(s.charAt(j++));
                int t = j - i;
                ans = (ans > t) ? ans : t;
            } else {
                set.remove(s.charAt(i++));
            }
        }
        return ans;
    }

官方答案3:两个字，厉害

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length(), ans = 0;
        Map<Character, Integer> map = new HashMap<>(); // current index of character
        // try to extend the range [i, j]
        for (int j = 0, i = 0; j < n; j++) {
            if (map.containsKey(s.charAt(j))) {
                i = Math.max(map.get(s.charAt(j)), i);
            }
            ans = Math.max(ans, j - i + 1);
            map.put(s.charAt(j), j + 1);
        }
        return ans;
    }
}