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
LeetCode 240. Search a 2D Matrix II 程序员木子 香港浸会大学 数据分析与人工智能硕士在读 来自专栏 · LeetCode 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 in ascending ...
principle:1. iff matrix[x][y] >target, since y is sorted, target must not able to appear on y column, we should skip it.while( y >= 1 && target <matrix[x][y] ) { y--; }2. iff matrix[x][y] <target. since matrix[x][y-1] must smaller than it, and matrix[x][y+1]...
* [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...
从左上角开始搜索。因为matrix[i][j]小于所有>i 或>j的元素,也就是讲只有0~i 和0~j的元素无法确定大小。 算法: public boolean searchMatrix(int[][] matrix, int target) { int j = matrix[0].length-1,i=0; while(i<matrix.length&&j>=0){ ...
2. 3. 4. 5. 6. 7. Given target = 5, return true. Given target = 20, return false. 题意:给定一个二维数组matrix和一个目标元素target,计算在二维数组中是否存在元素target;其中,数组matrix每一行都是有序的; 解法:虽然数组matrix中的每一行都是有序的,但是与Search a 2D matrix不同的是,不满足每...
这道题让我们在一个二维数组中快速的搜索的一个数字,这个二维数组各行各列都是按递增顺序排列的,是之前那道Search a 2D Matrix 搜索一个二维矩阵的延伸,那道题的不同在于每行的第一个数字比上一行的最后一个数字大,是一个整体蛇形递增的数组。所以那道题可以将二维数组展开成一个一位数组用一次二查搜索。而...
/* * @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: 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...
func searchMatrix(matrix [][]int, target int) bool { // 矩阵行数 m := len(matrix) // 矩阵列数 n := len(matrix[0]) // 二分区间左边界初始化为 0 l := 0 // 二分区间右边界初始化为 m * n - 1 r := m * n - 1 // 当二分区间至少还存在 2 个数时,继续二分 for l < ...