842. Split Array into Fibonacci Sequence —— weekly contest 86

原题链接在这里:https://leetcode.com/problems/split-array-into-fibonacci-sequence/ 题目: Given a string S of digits, such as S = "123456579", we can split it into a Fibonacci-like sequence [123, 456, 579]. Formally, a Fibonacci-like sequence is a list F o

Given a string S of digits, such as S = "123456579", we can split it into a Fibonacci-like sequence [123, 456, 579]. Formally, a Fibonacci-like sequence is a list F of non-negative integers such that: 0 <= F[i] <= 2^31 - 1, (that is, each

Weekly Contest 86 A:840. 矩阵中的幻方 3 x 3 的幻方是一个填充有从 1 到 9 的不同数字的 3 x 3 矩阵,其中每行,每列以及两条对角线上的各数之和都相等. 给定一个由整数组成的 N × N 矩阵,其中有多少个 3 × 3 的 "幻方" 子矩阵?(每个子矩阵都是连续的). 直接模拟即可,本来是签到题,由于粗心,浪费了时间. 1 class Solution { 2 public: 3 int numMagicSquaresInside(vector&l

题目链接:https://leetcode.com/problems/magic-squares-in-grid/description attention:注意给定的数字不一定是1-9. time:5ms 本人的解法过于粗糙,看出了中间必须是5,然后比较每行每列每对角线的值是否相等. class Solution { public: int numMagicSquaresInside(vector<vector<int>>& grid) { int n = grid.si

题目链接:https://leetcode.com/problems/keys-and-rooms/description/ 简单DFS time:9ms 1 class Solution { 2 public: 3 void DFS(int root,vector<int>& visited,vector<vector<int>>& rooms){ 4 visited[root] = 1; 5 for(auto x : rooms[root]){ 6

题目链接:https://leetcode.com/problems/guess-the-word/description/ 占坑 据说要用启发式算法,可参考下述答案进行学习:https://leetcode.com/problems/guess-the-word/discuss/133862/Random-Guess-and-Minimax-Guess-with-Comparison 原文地址:https://www.cnblogs.com/jinjin-2018/p/9098158.html

斐波那契数列(Fibonacci sequence)的T-SQL实现 ;WITH T AS ( SELECT 1 AS NUM, CAST(1 AS BIGINT) AS curr, CAST(NULL AS BIGINT) AS prv UNION ALL SELECT curr.NUM + 1 AS NUM, CAST(CASE WHEN prv IS NULL THEN curr ELSE curr + prv END AS BIGINT) AS curr, CAST(curr AS BI

1779. Fibonacci Sequence Multiplication Constraints Time Limit: 1 secs, Memory Limit: 63.9990234375 MB Description Maybe all of you are familiar with Fibonacci sequence. Now you are asked to solve a special version of Fibonacci sequence: The Multipli

使用Python实现斐波那契数列(Fibonacci sequence) 斐波那契数列形如 1,1,2,3,5,8,13,等等.也就是说,下一个值是序列中前两个值之和.写一个函数,给定N,返回第N个斐波那契数字.例如,1返回1 6返回8 我选择了两种方法,一种是将list变成一个队列,另一个则是使用环形队列.不多说,直接上代码:后面我会对为什么这样实现做一个解释 第一个是使用队列的方式: 1 def fibonacciSeq(num): 2 fibonacciSeqList = [] 3 for