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 from left to right. The first integer of each row is greater than the last integer of the previous ro…

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 左下或者右上开始查找 顺序查找 库函数 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 m…

Difficulty:medium  More:[目录]LeetCode Java实现 Description 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 from left to right. The first integer of each…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 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 from lef…

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 from left to right. The first integer of each row is greater than the last integer of the previous ro…

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…

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…

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…

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…