3248 Snake in Matrix 题意:给定n x n方阵,坐标位置(i,j)用i*n+j表示。机器人从0位置出发,按照给定的上下左右指令移动。请求出最终位置。 难度:easy 解法:水题。3249 Count the Number of Good Nodes 题意:给定一个树形的无向图,n个点编号0~n-1,根节点是0。如果某个节点的所有子树的节点数都相同
3533-snake-in-matrix 3534-count-almost-equal-pairs-i 3536-find-the-count-of-monotonic-pairs-ii 3540-hash-divided-string 3541-report-spam-message 3542-maximum-value-sum-by-placing-three-rooks-ii 3543-count-substrings-that-satisfy-k-constraint-i 3546-count-substrings-that-satisfy-k-constraint-...
3248 Snake in Matrix Easy Python 3249 Count the Number of Good Nodes Medium Python 3264 Final Array State After K Multiplication Operations I Easy Python 3370 Smallest Number With All Set Bits Easy Python 3371 Identify the Largest Outlier in an Array Medium Python 3372 Maximize the Number of Ta...
329 Longest Increasing Path in a Matrix // #329 矩阵最长递增路径 描述:给定一个矩阵,允许从某点出发,上下左右走。请找出能走出的单调递增序列的最大长度。 //#329Description: Longest Increasing Path in a Matrix | LeetCode OJ 解法1:可以搜,也可以DP,道理其实是一样的。 // Solution 1: BFS or DP...
4y1q7Lpleetcode 71 Simplfy Path BV1S54y1v7USleetcode 74 Search a 2D Matrix BV1aZ4y1u7weleetcode 79 Word Search BV175411p7y2leetcode 80 Remove Duplicates From Sorted Array II BV1PT4y1E78Aleetcode 82 Remove Duplicates from Sorted Lists II BV1TK4y1W7Celeetcode 84 Largest Rectangle in ...
LeetCode分类-前400题 1. Array# 基础 2. String# 基础 3. Math# 基础 4. Tree# 基础 5. Backtracking# 基础 6. Dynamic Programming# 一维 7. LinkedList# 基础 8. Binary Search# 基础 9. Matrix#
if(matrix.empty()) return false; int M = matrix.size(); int N = matrix[0].size(); int i = 0, j = N-1; while( i < M && j >=0){ if(matrix[i][j] == target) return true; else if(matrix[i][j] < target) i++; ...
截止至今LeetCode题目总量已经有1582题,估计将来每年平均增长300题左右,大部分人肯定是刷不完的,所以得有选择地刷LeetCode。 一种公认的刷题策略是按类别刷题,可是每个类别也有许多题,在有限的时间里到底该刷哪些题呢?个人根据LeetCode官方给出的每个题目的出现频率,整理并收录了每个类别里高频出现的题目,对于官方统...
Can you solve this real interview question? Minimum Moves to Reach Target with Rotations - In an n*n grid, there is a snake that spans 2 cells and starts moving from the top left corner at (0, 0) and (0, 1). The grid has empty cells represented by zeros
Follow up: Could you do this in-place? public void rotate(int[][] matrix) { int n = matrix.length; int first, last, os, temp; for (int i = 0; i n / 2; ++i) { first = i; last = n - i - 1; for (int j = first; j last; ++j) { os = j - i; temp = matrix...