Su**us上传2KB文件格式zip LongestStringDistinctCharacters 包含不同字符的最长子字符串 (0)踩踩(0) 所需:1积分 gaussnewton逻辑斯蒂克曲线 2025-01-11 19:46:20 积分:1 JqueryCombinedCommissioningPlugin 2025-01-11 19:45:40 积分:1 华为商城自动登录脚本,解决手动登录账号的麻烦 ...
再引入一个numDistinct变量,用于记录窗口中已经出现的不同字符次数。此外,使用一个int[258]来模拟哈希表,记录每个字符出现的次数count。 遍历字符串(right右移),当遇到的字符出现次数为0时,numDistinct++; 当numDistinct的次数超过2时,说明left需要往右移,在left往右移的过程中,减少所指字符的count,最终使得numDistinc...
welcome to my blog LeetCode Top Interview Questions 340. Longest Substring with At Most K Distinct Characters (Java版; Hard) 题目描述 Given a string, find the length of the longest substring T that contains at most k distinct characters. Example 1: Input: s = "eceba", k = ...
2. 出现两种char, 保证p1 <= p2, 为了计算方便 3. 出现第三种char的时候,i - (p1+1) + 1 = i- p1 直接计算出当前substring长度。同时保证p1<=p2. public class Solution { public int lengthOfLongestSubstringTwoDistinct(String s) { if(s == null || s.length() == 0) return 0; int p1...
Java public class Solution { public int lengthOfLongestSubstringTwoDistinct(String s) { String longestSubstr = ""; int maxLength = 0, start = 0; HashMap<Character, Integer> map = new HashMap<Character, Integer>(); for(int i = 0; i < s.length(); i++){ ...
类似Longest Substring with At Most Two Distinct Characters. 2变成k即可. Time Complexity: O(n), n = s.length(). Space: O(1), map size. AC Java: 1publicclassSolution {2publicintlengthOfLongestSubstringKDistinct(String s,intk) {3intres = 0;4int[] map =newint[256];5intwalker = 0;...
Given a string, find the length of the longest possible substring in it that has exactly K distinct characters. If there is no possible substring then print -1. You can assume that K is less than or equal to the length of the given string. Example 1: Input: S = "aabacbebebe", K ...
Input: String="cbbebi", K=3 Output: 5 Explanation: The longest substrings with no more than '3' distinct characters are "cbbeb" & "bbebi". 2.2 解决方法 该问题为含有 K 个不同字符的最长子串问题,首先可以想到可以运用暴力破解方法求得所有符合要求的字符串,然后再求得每个字符串的长度选择最长...
distinct characters. For example, Given s=“eceba”, T is"ece" which its length is 3. 1. 2. 3. 4. 5. 方法一:用HashMap, map里面存元素及其出现次数。维护一个最大长度。用两个指针,右指针一直走到出现3个dinstinct character为止。然后调整左指针删元素,直接从左往右逐个字符的删除,一直删到某个...
Java 已开启智能提示 30 分 00 秒 1 2 3 4 5 6 7 8 9 10 public·class·Solution·{ ···/** ···*·@param·s:·A·string ···*·@param·k:·An·integer ···*·@return:·An·integer ···*/ ···public·int·lengthOfLongestSubstringKDistinct(String·s,·int·k)·{...