Can you solve this real interview question? Search a 2D Matrix II - Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This matrix has the following properties: * Integers in each row are sorted in ascendin
y= len(matrix[0]) - 1forxinrange(len(matrix)):whileyandmatrix[x][y] >target: y-= 1ifmatrix[x][y] ==target:returnTruereturnFalse 解法二:循环枚举行,二分查找列 classSolution(object):defsearchMatrix(self, matrix, target):#二分查找 148ms""":type matrix: List[List[int]] :type targ...
* [240] Search a 2D Matrix II */// @lc code=startclassSolution{public:boolsearchMatrix(vector<vector<int>>& matrix,inttarget){ assert(!matrix.empty() && !matrix[0].empty());intm = matrix.size(), n = matrix[0].size();inti = m -1, j =0;while(i >=0&& j < n) {if(mat...
因为每一列都是增序的,举个例子,假设matrix[0][5] > target,那么[0][5]位置右下(包含右和下)所有元素不可能比target小。 题目链接: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 ro...
LeetCode_240. Search a 2D Matrix II 题目描述: 思路:题目要求写一个高效率的算法,所以暴力遍历整个矩阵查找 的方法行不通了。 分析题目,发现这个矩阵是从左到右,从上到下递增的。可以发现矩阵右上角的元素是一个特殊的元素( ):如果 和 正好相等了,说明找到了。而当...
leetcode 240. Search a 2D Matrix II 矩阵搜索 + 右上角搜索,Writeanefficientalgorithmthatsearchesforavalueinanmxnmatrix.Thismatrixhasthefollowingproperties:Integersineachrowaresortedinach
* 题目: 240.Search a 2D Matrix II * 网址:https://leetcode.com/problems/search-a-2d-matrix-ii/ * 结果:AC * 来源:LeetCode * 博客: ---*/#include<iostream>#include<vector>#include<stack>usingnamespacestd;classSolution{public:boolsearchMatrix(vector<vector<int>>& matrix,inttarget){if(mat...
/* * @lc app=leetcode id=240 lang=javascript * * [240] Search a 2D Matrix II * * https://leetcode.com/problems/search-a-2d-matrix-ii/description/ * * *//** * @param {number[][]} matrix * @param {number} target * @return {boolean} */var searchMatrix = function (matrix, ...
题目: Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 Integers in each row are sorted from left to right. The first integer of each row is greater than the last integer...
length-1; while(left<right){ int mid=(left+right)/2; if(matrix[row][mid]==target) return true; if(matrix[row][mid]>target) right=mid-1; else left=mid+1; } return matrix[row][left]==target; } public boolean searchMatrix(int[][] matrix, int target) { int m=matrix.length; ...