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…

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.…

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…

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…

二分查找 1.二分查找的时间复杂度分析: 二分查找每次排除掉一半不合适的值,所以对于n个元素的情况来说: 一次二分剩下:n/2 两次:n/4 m次:n/(2^m) 最坏情况是排除到最后一个值之后得到结果,所以:n/(2^m) = 1 2^m = n 所以时间复杂度为:log2(n) 2.二分查找的实现方法: (1)递归 int RecursiveBinSearch(int arr[], int bottom, int top, int key) { if (bottom <= top) { in…

题目如下:这两个题目可以用同样的代码来解答,因此就合并在一起了. 题目一: 题目二: 解题思路:两个题目的唯一区别在于第二个题目下一行的最小值不一定会小于前一行的最大值.但是不管怎么样我们可以确定的是,如果某一行的最小值都比target要大,那么这一行之后的值都比target要大.如果target介于某一行的最小值和最大值之间,那么target有可能在这一行.至于如何判断target是否存在,因为数组有序,用二分查找即可. 代码如下: class Solution(object): def se…