Given a string, find the length of the longest substring without repeating characters. 找出字串S中最长无重复子串的长度。 Examples: Given"abcabcbb", the answer is"abc", which the length is 3. Given"bbbbb", the answer is"b", with the length of 1. Given"pwwkew", the answer is"wke",...
classSolution {public:intlongestSubstring(strings,intk) {intres =0, n =s.size();for(intcnt =1; cnt <=26; ++cnt) {intstart =0, i =0, uniqueCnt =0; vector<int> charCnt(26);while(i <n) {boolvalid =true;if(charCnt[s[i++] -'a']++ ==0) ++uniqueCnt;while(uniqueCnt >cnt...
publicclassSolution{publicintlengthOfLongestSubstring(Strings){intn=s.length();intans=0;for(inti=0;i<n;i++)for(intj=i+1;j<=n;j++)if(allUnique(s,i,j))ans=Math.max(ans,j-i);returnans;}publicbooleanallUnique(Strings,intstart,intend){Set<Character>set=newHashSet<>();for(inti=start...
publicclassSolution{publicintlengthOfLongestSubstring(Strings){intn=s.length();intans=0;for(inti=0;i<n;i++)for(intj=i+1;j<=n;j++)if(allUnique(s,i,j))ans=Math.max(ans,j-i);returnans;}publicbooleanallUnique(Strings,intstart,intend){Set<Character>set=newHashSet<>();for(inti=start...
Given a string you need to print the size of the longest possible substring that has exactly k unique characters. If there is no possible substring print -1. Example For the string aabacbebebe and k = 3 the substring will be cbebebe with length 7. ...
We also need a way to check whether this window has repeating letters - hash map is a better way, but how to construct the hash map and how this map functions? Solutions: When we slide the window, the hash map is constructed with " key of unique letter " and " value of number " ...
Suppose we have a functionboolean allUnique(String substring)which will return true if the characters in the substring are all unique, otherwise false. We can iterate through all the possible substrings of the given stringsand call the functionallUnique. If it turns out to be true, then we ...
publicclassSolution{publicintlengthOfLongestSubstring(String s){intn=s.length();intans=0;for(inti=0; i < n; i++)for(intj=i +1; j <= n; j++)if(allUnique(s, i, j)) ans = Math.max(ans, j - i);returnans; }publicbooleanallUnique(String s,intstart,intend){ ...
the answer is: when the substring is totally random. it is not possible, but if we can make sure that head and tail of this sliding window are unique all the time, the sliding window will always be legal? the answer is no. just use abb as example with initial pointer on a and the...
importjava.util.Scanner;importjava.util.HashMap;importjava.util.Map;classLongestSubstringWithoutRepeatingCharacters{// Variant of Longest Substring with K unique characters// Here it is Longest substring with ALL unique characters// (So compare against the whole window length windowEnd-windowStart+1 ...