Given a stringS, find the longest palindromic substring inS. You may assume that the maximum length ofSis 1000, and there exists one unique longest palindromic substring. 求字符串的最长回文子串 算法1:暴力解法,枚举所有子串,对每个子串判断是否为回文,复杂度为O(n^3) 算法2:删除暴力解法中有很多重复...
1publicString longestPalindrome(String s) {2intn =s.length();3String ans = "";4Set<String> set =newHashSet<String>();5for(inti = 0; i < n; i++) {6for(intj = i + 1; j <= n; j++) {7set.add(s.substring(i, j));8}9}10for(String m : set) {11if(isPalindromic(m...
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters. For example, given: s: “barfoothefoobarman” wo...
Return the starting indices of all the concatenated substrings ins. You can return the answer inany order. usestd::collections::HashMap;implSolution{pubfnfind_substring(s:String,words:Vec<String>)->Vec<i32>{letwlen=words[0].len();lettotal=s.len();letword_count=words.len();letmutret=V...
classSolution{publicintlengthOfLongestSubstring(Strings){if(s.equals("")){return0;}else{intcount;intecount=0;for(inti=0;i<s.length();i++){count=1;for(intj=i+1;j<s.length();j++){if(s.substring(i,j).contains(String.valueOf(s.charAt(j))){break;}count++;}if(count>ecount){ecoun...
toBinaryString(i | nthBit).substring(1); 或者使用简单但低效的迭代进行控制: Python 实现 for i in range(2**n, 2**(n + 1)): # generate bitmask, from 0..00 to 1..11 bitmask = bin(i)[3:] Java 实现 for (int i = (int)Math.pow(2, n); i < (int)Math.pow(2, n + 1)...
public int longestSubstring(String s, int k) { if (s.length() < k) return 0; HashMap<Character, Integer> counter = new HashMap(); for (int i = 0; i < s.length(); i++) { counter.put(s.charAt(i), counter.getOrDefault(s.charAt(i), 0) + 1); ...
classSolution{public:intlengthOfLongestSubstring(string s){if(s.length()==0){return0;}unordered_map<char,int>hash;intbegin=0,end=0;intrepeat=0;intmax_length=0;while(end<s.length()){if(hash[s[end]]>0){repeat++;}hash[s[end]]++;end++;while(repeat>0){if(hash[s[begin]]>1){repe...
public int numUniqueEmails(String[] emails) { Set<String> emailSet = new HashSet<String>(); for (String email : emails) { int i = email.indexOf('@'); String local = email.substring(0, i).split("\\+")[0]; // 去掉本地名第一个加号之后的部分 ...
https://leetcode-cn.com/problems/minimum-window-substring/ Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). 题意 给你一个字符串 S、一个字符串 T 。请你设计一种算法,可以在 O(n) 的时间复杂度内,从字符串 ...