classSolution{publicintmaximumLength(int[]nums){int res=0,k=2,dp[][]=newint[k][k];for(int a:nums){for(int b=0;b<k;b++){dp[b][a%k]=Math.max(dp[b][a%k],dp[a%k][b]+1);res=Math.max(res,dp[b][a%k]);}}returnres;}} 还是不熟悉dp。。。
String t is generated by random shuffling string s and then add one more letter at a random position. Return the letter that was added to t. Example 1: Input: s = "abcd", t = "abcde" Output: "e" Explanation: 'e' is the letter that was added. Example 2: Input: s = "", ...
}classSolution {public:intmaxPoints(vector<Point> &points) {intmax=0;for(vector<Point>::iterator it=points.begin(); it!=points.end(); ++it) { map<double,int>Line_count;intsame_count=0; //same_count表示同一个点重复出现的个数for(vector<Point>::iterator it2=it; it2!=points.end()...
438. Find All Anagrams in a String # 题目 # Given a string s and a non-empty string p, find all the start indices of p’s anagrams in s. Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 2
18 changes: 18 additions & 0 deletions 18 2769.find-the-maximum-achievable-number.py Original file line numberDiff line numberDiff line change @@ -0,0 +1,18 @@ # # @lc app=leetcode id=2769 lang=python # # [2769] Find the Maximum Achievable Number # # @lc code=start class Solut...
Can you solve this real interview question? Find Maximum Non-decreasing Array Length - You are given a 0-indexed integer array nums. You can perform any number of operations, where each operation involves selecting a subarray of the array and replacing
For array of 1 element, there is 1 way for each different number choice with a search cost of 1. This is true because maximum value is initially set to < 0, so any number choice will incur an update. dp[0][1][maxV] = 1, maxV in [1, m]. ...
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给你一个下标从0开始的整数数组nums。 一开始,所有下标都没有被标记。你可以执行以下操作任意次: 选择两个互不相同且未标记的下标i和j,满足2 * nums[i] <= nums[j],标记下标i和j。 请你执行上述操作任意次,返回nums中最多可以标记的下标数目。