8/16 每日一題(找出最長不連續子序列)

 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.

class Solution:

    def lengthOfLongestSubstring(self, s: str) -> int:

        if len(s)==1:

            return 1

        if s=='':

            return 0

        n=0

        sub_n=1

        max_len=1

        res=set()

        while n < len(s):

            if s[n] in res:

                #遇到相同時,清算

                max_len=max(len(res),max_len)

                res.clear() #清算

                #清算後由下一項開始檢查

                n=sub_n

                sub_n+=1

                continue#回到起始

            else:

                res.add(s[n])

                n+=1

        max_len=max(len(res),max_len)

        return max_len

-------------------------參考解答

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        seen = {}
        l = 0
        length = 0
        for r in range(len(s)):
            char = s[r]
            if char in seen and seen[char] >= l:
                l = seen[char] + 1
            else:
                length = max(length, r - l + 1)
            seen[char] = r

        return length

-----------------------

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        l, r = 0, 0
        res = 0
        cs = set()
        while r < len(s):
            if s[r] in cs:
                while s[l] != s[r]:
                    cs.remove(s[l])
                    l += 1
                l += 1
            else:
                cs.add(s[r])
            res = max(res, r - l + 1)
            r += 1
        return res