[编程练习][Longest Substring Without Repeating Characters]<LeetCode-3>

题目来自https://oj.leetcode.com/problems/longest-substring-without-repeating-characters/

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.

我的解答：

class Solution:    # @return an integer    def lengthOfLongestSubstring(self, s):        start, curlen, findlen = 0,0, 0        Seen = dict()        lenstr = len(s)                for i in range(lenstr):            ch = s[i]            if ch not in Seen or Seen[ch] < start:                curlen += 1            else:                if curlen > findlen:                    findlen = curlen                start = Seen[ch] + 1                curlen = i - Seen[ch]            Seen[ch] = i        #if found at end        if curlen > findlen:            findlen = curlen                    return findlen

说明：

这一题的难度标注为“Medium"， 算是中规中矩吧。

我的思路是构造字典，记录出现过的字符的位置。从头开始扫描，统计无重复字符串的长度。

当前字符ch没有重复出现，则更新当前子串的无重复长度。start 指的是新串起始位置。

            if ch not in Seen or Seen[ch] < start:                curlen += 1

重复出现，则检查是否是当前最大解，是则更新。记录新串起始位置和新串长度（利用Seen[ch]

            else:                if curlen > findlen:                    findlen = curlen                start = Seen[ch] + 1                curlen = i - Seen[ch]            Seen[ch] = i

程序很简单，重点是利用Seen[ch]来生成每次重新开始新串的起始位置和既有长度。

Python 的字典又一次发挥了作用。