D1.Remove the Substring (easy version) time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output The only difference between easy and hard versions is the length of the string. You are given a string…
D2. Remove the Substring (hard version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The only difference between easy and hard versions is the length of the string. You are given a stri…
D2. Remove the Substring (hard version) 思路:其实就是贪心吧,先从前往后找,找到 t 可在 s 中存在的最小位置 (pre),再从后往前找,找到 t 可在 s 中存在的最大位置(last),然后 last [ i+1 ] - pre [ i ] - 1 表示的即是 t 中两个相邻字符可以删去的最大连续字串长度,当然也不要忘记第一个字符前面删去( last [ 0 ] )的和最后末尾删去的字符串(slen - pre [ tlen - 1 ] - 1) 代码…
DescriptionDr lee cuts a string S into N pieces,s[1],…,s[N]. Now, Dr lee gives you these N sub-strings: s[1],…s[N]. There might be several possibilities that the string S could be. For example, if Dr. lee gives you three sub-strings {“a”,“ab”,”ac”},…
题目链接 Problem Description When online chatting, we can save what somebody said to form his ''Classic Quotation''. Little Q does this, too. What's more? He even changes the original words. Formally, we can assume what somebody said as a string S whose…
Given a pattern and a string str, find if strfollows the same pattern. Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty substring in str. Example 1: Input: pattern = "abab", str = "r…
Given a list of words, we may encode it by writing a reference string S and a list of indexes A. For example, if the list of words is ["time", "me", "bell"], we can write it as S = "time#bell#" and indexes = [0, 2,…
class Solution { public List<String> removeInvalidParentheses(String s) { List<String> ans = new ArrayList<>(); remove(s, ans, 0, 0, new char[]{'(', ')'}); return ans; } public void remove(String s, List<String> ans, int last_i, in…
2018-07-02 09:48:48 问题描述: 问题求解: 方法一.问题给了规模n = 2000,也就是说用BF在O(n^2)的时间复杂度可以过,因此,第一个方法就是BF,但是需要注意的是这里已经非常擦边了,所以需要对常数项进行优化. public int minimumLengthEncoding(String[] words) { boolean[] flag = new boolean[words.length]; int[] lens = new int[words.length];…
Classic Quotation Problem Description When online chatting, we can save what somebody said to form his ''Classic Quotation''. Little Q does this, too. What's more? He even changes the original words. Formally, we can assume what somebody said as a st…
Classic Quotation Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others) Problem Description When online chatting, we can save what somebody said to form his ''Classic Quotation''. Little Q does this, too. What's more?…
Codeforces Round #579 (Div. 3) 传送门 A. Circle of Students 这题我是直接把正序.逆序的两种放在数组里面直接判断. Code #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 205; int q, n; int a[N], b[N], c[N]; int main() { ios::sync_with_stdio(false);…
A. Circle of Students 题目:https://codeforces.com/contest/1203/problem/A 题意:一堆人坐成一个环,问能否按逆时针或者顺时针正好是 1-n的顺序 思路:水题,把数组开两倍,或者标记当前位置都可以 #include<bits/stdc++.h> #define maxn 100005 #define mod 1000000007 using namespace std; typedef long long ll; int…
比赛链接:https://codeforc.es/contest/1203/ A. Circle of Students 题意:\(T\)组询问,每组询问给出\(n\)个数字,问这\(n\)个数字能否围成圆环.(围成圆环指,从某一位开始顺时针或逆时针遍历,数组为\(1, 2, 3, ..., n\)) 分析:把数组复制一份,两个数组首尾相接,正反判定两次即可. AC代码: #include<bits/stdc++.h> #define SIZE 200010 #define rep(i, a,…
Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest subst…
