https://leetcode.com/problems/longest-substring-without-repeating-characters/
Given a string, find the length of the longest substring without repeating characters.
Example 1:
Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc"
, with the length of 3.
Example 2:
Input: "bbbbb"
Output: 1
Explanation: The answer is "b"
, with the length of 1.
Example 3:
Input: "pwwkew"
Output: 3
Explanation: 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.
- 字符串简单题。Sliding window + set / dict
- 基础算法是把把char都放到一个set里,left指针按照步长为1往右移动并判重。
- 优化算法是出现重复时,直接把window右移到不重复位置再继续,即left指针移动到重复位置+1。
- https://leetcode.com/problems/longest-substring-without-repeating-characters/solution/
- The above solution requires at most 2n steps. In fact, it could be optimized to require only n steps. Instead of using a set to tell if a character exists or not, we could define a mapping of the characters to its index. Then we can skip the characters immediately when we found a repeated character.
- The reason is that if s[j] have a duplicate in the range [i, j) with index j', we don't need to increase i little by little. We can skip all the elements in the range [i, j'] and let i to be j' + 1 directly.
- Python3 字典 | 菜鸟教程
- http://www.runoob.com/python3/python3-dictionary.html
- Python3 集合 | 菜鸟教程
- http://www.runoob.com/python3/python3-set.html
1 class Solution: 2 def lengthOfLongestSubstring(self, s: str) -> int: 3 d = dict() # current index of character 4 left, right, maxlen = 0, 0, 0 5 n = len(s) 6 7 while right < n: 8 # if char in dict[j'],skip all the elements in the range [i,j'] and let i to be j'+1 directly. 9 if s[ right ] in d: 10 left = max( d[ s[ right ] ], left ) 11 12 maxlen = max( right - left + 1, maxlen ) 13 14 # set next available index for left, i.e. j'+1 15 d[ s[ right ] ] = right + 1 16 17 right += 1 18 19 return maxlen 20 21 def lengthOfLongestSubstring1(self, s: str) -> int: 22 dict = set() 23 24 n = len(s) 25 left, right, maxlen = 0, 0, 0 26 27 while right < n: 28 if s[ right ] not in dict: 29 dict.add( s[ right ] ) 30 maxlen = max( maxlen, right - left + 1 ) 31 right += 1 32 else: 33 dict.remove( s[ left ] ) 34 left += 1 35 36 return maxlen