give a string, all 1 or 0, we can flip a 0 to 1, find the longest 1 substring after the flipping 这是一个简单版本of LC 424 Longest Repeating Character Replacement 又是Window, 又是Two Pointers Window还是采用每次都try to update left使得window valid, 每次都检查最大window package G…

原题链接在这里:https://leetcode.com/problems/swap-for-longest-repeated-character-substring/ 题目: Given a string text, we are allowed to swap two of the characters in the string. Find the length of the longest substring with repeated characters. Example 1: In…

题目如下: Given a string text, we are allowed to swap two of the characters in the string. Find the length of the longest substring with repeated characters. Example 1: Input: text = "ababa" Output: 3 Explanation: We can swap the first 'b' with the…

Given two strings, find the longest common subsequence (LCS). Your code should return the length of LCS. Example For "ABCD" and "EDCA", the LCS is "A" (or "D", "C"), return 1. For "ABCD" and &quo…

Given a string that consists of only uppercase English letters, you can replace any letter in the string with another letter at most k times. Find the length of a longest substring containing all repeating letters you can get after performing the abo…

DP use HashMap: 根据string的长度sort,然后维护每个string的longest chain,default为1,如果删除某个char生成的string能提供更长的chain,则更新 package twoSigma; import java.util.Arrays; import java.util.Comparator; import java.util.HashMap; import java.lang.StringBuilder; public class Lon…

求string str1中含有string str2 order的 subsequence 的最小长度 DP做法:dp[i][j]定义为pattern对应到i位置,string对应到j位置时,shortest substring的长度,Int_Max表示不存在 package ShortestSubsequenceIncluding; public class Solution { public String findShortest(String a, String b){ if(a==nul…

求两个sorted数组的intersection e.g. [1,2,3,4,5],[2,4,6] 结果是[2,4] difference 类似merge, 分小于等于大于三种情况,然后时间O(m+n), 空间O(1) package ShortestSubsequenceIncluding; import java.util.*; public class Solution2 { public ArrayList<Integer> setInters(int[] arr1, int[] ar…

字符串重新排列,让里面不能有相同字母在一起.比如aaabbb非法的,要让它变成ababab.给一种即可 Greedy: 跟FB面经Prepare task Schedule II很像,记录每个char出现次数,然后用最大堆,把剩下char里面出现次数多的优先Poll出来组建新的string 如果poll出来的char跟上一个相同,则用一个queue暂时存一下 我觉得时间复杂度:O(N) + O(KlogK) + O(NlogK) = O(NlogK) ,where K is the number…

给定一个字符串,找出最多有多少个chunked palindrome, 正常的palindrome是abccba, chunked palindrome的定义是:比如volvo, 可以把vo划分在一起,(vo) (l) (vo),那么它是个palindrome.求实现返回最大的chunk 数量. 比如aaaaaa可以是(aaa)(aaa), 但是最大chunk数量应该是(a)(a)(a)(a)(a)(a)为6 这就是一个greedy的问题,从string的两边开始,用i和j记录当前scan到的位…