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3. Longest Substring Repeating Character

Difficulty:Medium

The link:

• https://leetcode.com/problems/longest-substring-without-repeating-characters/

Description :

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

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.

Solutions:

Solution A:

(* 为解决) 想实现KMP算法 　KMP 介绍

class Solution:
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        look_up = {}
        for i in range(len(s)):
            if s[i] not in look_up:
                look_up[s[i]] = i
            else:
                # l 现在字典的长度
                l = len(look_up)
                # m 出现重复字母位置
                m = look_up[s[i]]
                # str01  字符串切片
                str01 = s[m:l]
                for i, item in enumerate(str01):
                    if s[i] not in look_up:
                        i = m + l
                        l = l + 1        
        return l

Solution B:

dict,get() The method get() returns a value for the given key. If key is not available then returns default value None.

dict.get(key, default = None)

• key - This is the Key to be searched in the dictionary.
• default - This is the Value to be returned in case key does not exist.

class Solution:
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        l, start, n = 0, 0, len(s)
        maps = {}
        for i in range(n):
            start = max(start, maps.get(s[i], -1)+1)
            l = max(l, i - start+1)
            maps[s[i]] = i
        return l

leetcode 003

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原文地址：https://www.cnblogs.com/Mr0wang/p/9721570.html