算法1

(双指针) $O(n)$

• i、j 维护相关不重复的子串
• 比如字符串“abcaaabc”

C++ 代码

class Solution {
public:
 int longestSubstringWithoutDuplication(string s) {
        int i = 0, j = 0;
        int res = 0;
        unordered_map<char, int>map;
        while (j < s.size()){           
            if (++map[s[j]] > 1 )
            {
           while(map[s[i]] == 1) map[s[i++]]--;//解决连续两个黏在一起，则需要重新开始
               map[s[i++]]--;
            }

            res = max(res,j - i + 1);
            j++;
        }

        return res;
    }
};

算法2

(动态规划)

动态规划，用f(i)表示以i个字符结尾不包含重复子字符串的最长长度，从左向右扫描

1、若第i个字符在之前没出现过，则 f(i) = f(i-1) + 1;

2、若第i个字符在之前出现过，

计算第i个字符距离上次出现之间的距离为d

(a)若d <= f(i-1)，则说明第i个字符上次出现在f(i-1)对应的不重复字符串之内，那么这时候更新 f(i) = d

(b)若d > f(i-1)，则无影响,f(i) = f(i-1) + 1

C++ 代码

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        if (s.length() == 0)
            return 0;
        int curLen = 0, maxLen = 0;
        int* position = new int[26];
        for (int i = 0; i < 26; i++)
            position[i] = -1;

        for (int i = 0; i < s.length(); i++){
            int preIndex = position[s[i] - 'a'];
            if (preIndex < 0 || i - preIndex > curLen) // 没出现过，或者d>f(i-1)
                curLen++;
            else{                      // 出现过了
                if (curLen > maxLen)
                    maxLen = curLen;
                curLen = i - preIndex  // f(i) = d
            }
            position[s[i] - 'a'] = i;
        }
        if (curLen > maxLen)
            maxLen = curLen;

        delete[] position;
        return maxLen;
};