Title: Given a string, you find out which does not contain a sub-length of the longest string of repeated characters.
Example 1:
Input: "abcabcbb"
Output: 3
Explanation: Because the longest substring of characters without repetition is "abc", so its length is 3.
Example 2:
Example 2:
Input: "bbbbb"
Output: 1
Explanation: Because the longest substring is repeated characters without "b", so that its length is 1.
Example 3:
Example 3:
Input: "pwwkew"
Output: 3
Explanation: Because the longest sub-string is repeated characters without "wke", so its length is 3.
Please note that your answer must be a substring of length, "pwke" is a sub-sequence, not a substring.
Ideas: the use of a sliding window and begins to view rearwardly from the first character, when the character is repeated to find k, the previous string k (including k) discarding all the characters, and then update maxlen size, until moving to the right.
python code implementation for details attention :( length)
def lengthOfLongestSubstring(s: str) -> int:
i=1 #当前字符的位置
begin=0
maxlen=0
dropedlen=0#记录舍去的字符串的长度
if len(s)==0:
return 0
if len(s)==1:
return 1
for i in range(len(s)):
if s[i] in s[begin:i]:
# 先更新maxlen
templ=len(s[begin:i])
print("s[i]="+s[i])
print("字串="+s[begin:i])
print("templ = "+str(templ))
maxlen=templ if(templ>maxlen) else maxlen
# 舍去字串s[begin:i]中s[i]字符及前面的字符
# 直接用s[begin:i].index(s[i])获得的是相对位置，不是在s中的位置，应该加上dropedlen
relativelen=s[begin:i].index(s[i])#重复的字符串的相对位置
begin=relativelen+dropedlen+1
dropedlen=begin#更新删除的长度
# 到字符串末再更新一次maxlen
templ=len(s[begin:i])+1
maxlen=templ if(templ>maxlen) else maxlen
return maxlen
s = "au"
print("答案："+str(lengthOfLongestSubstring(s)))