Leetcode 3. Longest Substring Without Repeating Characters (Medium)

Description

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

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

Solution

Approach 1 :

用start, end 标记当前判别的子序列的起始与终止index。
1. 若当前序列s[start: end + 1] 没有重复字符，更新Max, 并 end + 1
2. 若当前新加入的s[end]与之前的子串subs中某元素重复，则定位subs中重复元素的index, 更新start为该index + 1，并更新subs.
依此规则，每一次end后移(+1)，则进行一次判断并更新。
最终Max即为所求最长无重复子串长度。

Notice:

subs = list(s[start : end + 1])
这里若直接subs = s[start : end + 1]，则subs为str类型，
需要转为list类型，即将str按字母元素分隔开。
eg. “abc” -> 'a','b','c'

因为subs中即为无重复子串，则无需用set,
可直接通过判断新加入的s[end]是否在subs中，来判断是否有重复字符

 1 class Solution:
 2     def lengthOfLongestSubstring(self, s):
 3         """
 4         :type s: str
 5         :rtype: int
 6         """
 7         
 8         if not s: return 0
 9         start, end, Max = 0, 0, 0
10         subs = []
11         while end < len(s):
12             if s[end] not in subs:
13                 subs.append(s[end])                          
14             else:
15                 start += subs.index(s[end]) + 1
16                 subs = list(s[start : end + 1])
17             Max = end - start + 1 if end - start + 1 > Max else Max
18             end += 1
19         return Max

Beats：35.65%
Runtime: 124ms

Approach 2:

基本思路与上面一致

 1 class Solution:
 2     def lengthOfLongestSubstring(self, s):
 3         """
 4         :type s: str
 5         :rtype: int
 6         """
 7         
 8         if not s:
 9             return 0
10         longestlength = 1
11         substr = ""
12         for item in s:
13             if item not in substr:
14                 substr += item
15             else:
16                 if len(substr) > longestlength:
17                     longestlength = len(substr)
18              
19                 #将重复字符加入substr
20                 substr += item
21                 # 应该从substr中第一次出现重复字符的下一个字符开始继续判断
22                 substr = substr[substr.index(item) + 1:]
23         
24         if len(substr) > longestlength:
25             longestlength = len(substr)
26         return longestlength

Beats: 44.09%
Runtime: 108ms

Approach 3: 用dict来存储判断重复元素

这里复制了leetcode上最快的答案。

解法中用到了

d: dict
i: 字母元素 d[i]: 更新后的字母元素在s中的index

每当遍历新的s元素，更新该元素在d中的映射值 (新元素：添加；已有元素：更新)
这样可以直接定位到其index, 而不需要每次通过s.index()查找

index表示当前遍历到的s中元素的ind
start表示当前子串的起始ind
count表示当前子串的长度

class Solution:
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        
        d = {}
        maxlength = 0
        count = 0
        start = 0
        for index, i in enumerate(s):
            if i in d and d[i] >= start:
                count = index - d[i]
                start = d[i]
            else:
                count+=1
            d[i] = index
            if count>maxlength:
                maxlength = count
            
        return maxlength

Beats: 99.93%
Runtime: 68ms