st当前无重复子串的最左边字符的位置 class Solution(): def lengthOfLongestSubstring_1(self, s): if s == None or len(s) <= 0: return charDict, res, st = {},0,0 for i, ch in enumerate(s): if ch not in charDict or charDict[ch] < st: res = max(res, i -st +1) else: ...
*@param{string}s*@return{number} */constlengthOfLongestSubstring =function(str) {letlen = str.length;if(len <1)return0;letres = str[0];letnewLen =1;letnewRes = str[0];for(leti =1; i < len; i++) {letj = res.indexOf(str[i]);if(j === -1) { res = res + str[i];...
解释: 因为无重复字符的最长子串是 "abc",所以其长度为 3。 classSolution:deflengthOfLongestSubstring(self,s:str)->int:ifnot s:return0words=[]count=len(s)foriinrange(count):word=""forjinrange(i,count):ifs[j]notinword:word+=s[j]else:breakwords.append(len(word))returnmax(words)if__nam...
复制 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(j))...
*/constlengthOfLongestSubstring =function(str) {letlen = str.length;if(len <1)return0;letres = str[0];letnewLen =1;letnewRes = str[0];for(leti =1; i < len; i++) {letj = res.indexOf(str[i]);if(j === -1) {
**lengthOflongestSubstring ** 试例: 输入: "abcabcbb" 输出: 3 解释: 因为无重复字符的最长子串是 "abc",所以其长度为 3。 /** * *@param{string}s* 无重复字符的最长子串 */functionlengthOfLongestSubstring(s) {if(typeofs !=='string'|| s.length===0)return0;constlen = s.length;letmax...
class Solution: def lengthOfLongestSubstring(self , s: str) -> int: # write code here # 方法一:滑动窗口 if not s: return 0 left = right = max_len = 0 while right < len(s): while s[right] in s[left: right]: left += 1 max_len = max(max_len, right - left + 1) right...
在下文中一共展示了Solution.lengthOfLongestSubstring方法的1個代碼示例,這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚,您的評價將有助於係統推薦出更棒的Python代碼示例。 示例1: main ▲點讚 7▼ # 需要導入模塊: from solve import Solution [as 別名]# 或者: from solve.Solution ...
LengthOfLongestSubstring.zipJo**ny 上传1.22 KB 文件格式 zip LengthOfLongestSubstring函数用于计算给定字符串中最长的不重复子串的长度。它通过维护一个滑动窗口来实现,窗口内的字符不重复。首先,定义两个指针,分别指向窗口的起始和结束位置。然后,遍历字符串,将字符加入窗口,如果字符已经在窗口中,则将起始指针向右...
public int lengthOfLongestSubstring(String s) { //使用滑动窗口 int maxLen = 0; Set<Character> set = new HashSet<>(); int rk = 0; int i; for(i=0;i