来自专栏 · python算法题笔记 class Solution: def longestContinuousSubstring(self, s: str) -> int: sub_len = 1 max_len = 1 for i in range(1, len(s)): if ord(s[i]) - ord(s[i-1]) == 1: sub_len +=1 max_len = max(max_len, sub_len) else: sub_len = 1 return max_...
Given a strings, find the length of the longest substring without repeating characters. Example 1: Input: s = "abcabcbb" Output: 3 Explanation: The answer is "abc", with the length of 3. Example 2: Input: s = "bbbbb" Output: 1 Explanation: The answer is "b", with the length of...
AI代码解释 publicclassL_00003_LengthOfLongestSubstring{publicintlengthOfLongestSubstring1(String s){if(s==null){return0;}Set<Character>charSet=newHashSet<>();int result=0;int len=s.length();for(int i=0;i<len;i++){int curLen=0;for(int j=i;j<len;j++){if(charSet.contains(s.charAt...
代码(Python3) class Solution: def longestContinuousSubstring(self, s: str) -> int: # ans 维护最长的字母序连续字符串的长度 ans: int = 0 # cnt 表示当前字母序连续字符串的长度, # 初始为字母序连续字符串仅由第一个字母组成,长度为 1 cnt: int = 1 # 遍历 s 中的每个字母 for i in range(...
# Example 1: Maximum length of string value maximum_length = sys.maxsize # Example 2: Get maximum length of string value # using max() and len() mystring = ["Spark","Python","Hadoop","Hyperion"] max_length = max(len(string) for string in mystring) ...
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Code Clone HTTPSGitHub CLI Download ZIP Latest commit Git stats 2commits Failed to load latest commit information. Description Given a string, find the length of the longest substring without repeating characters. Example Given "abcabcbb", the answer is "abc", which the length is 3. Given "...
Learn how to find the length of the longest set of 1s in a binary array by flipping K bits using Python programming.
public String longestCommonPrefix(String[] strs) { if(strs.length == 0) return ""; String prefix = strs[0]; for(int i = 1; i < strs.length; i++){ while(strs[i].indexOf(prefix) != 0){ prefix = prefix.substring(0, prefix.length() - 1); ...
{//根据ID获得指定标签的内容vartag=document.getElementById('i1')varcontent=tag.innerText;varf=content.charAt(0);//获取第一个字varl=content.substring(1, content.length);//获取第二个字到最后varnew_content=l+f;//重新组合tag.innerText=new_content;//重新赋值} setInterval('func()',500);//...