240. Search a 2D Matrix II】的更多相关文章

74. Search a 2D Matrix 整个二维数组是有序排列的,可以把这个想象成一个有序的一维数组,然后用二分找中间值就好了. 这个时候需要将全部的长度转换为相应的坐标,/col获得x坐标,%col获得y坐标 class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { int row = matrix.size(); ) return false; ].s…
Search a 2D Matrix II Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascendin…
Search a 2D Matrix II Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascendin…
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascending from top to bottom.…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/search-a-2d-matrix/description/ 题目描述 Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the follow…
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascending from top to bottom.…
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascending from top to bottom.…
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascending from top to bottom.…
1.问题描写叙述 写一个高效的算法.从一个m×n的整数矩阵中查找出给定的值,矩阵具有例如以下特点: 每一行从左到右递增. 每一列从上到下递增. 2. 方法与思路 2.1 二分查找法 依据矩阵的特征非常easy想到二分法,可是这是一个二维的矩阵,怎样将问题转化为一维是关键.实际上我们能够依据矩阵的第一列确定值可能所在的行的范围(limu,limd),当中limu=0,使得matrix[0][0]≤matrix[i][0]≤matrix[limd][0],i∈[0,limd]. 而确定limd的值能…
一句话思路:从左下角开始找.复杂度是o(m+n). 一刷报错: 应该是一个while循环中有几个条件判断语句,而不是每个条件判断语句里去加while,很麻烦 数组越界了.从0开始,y的最大条件应该是<n,不是<n 风格上:++ -- 都是一个数字的一元运算,不用加空格 超时了.循环体结构: while (x >= 0 && y <= n) { if (matrix[x][y] == target) { count ++; } else if (matrix[x][y…