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 ...
题目链接:https://leetcode.com/problems/search-a-2d-matrix-ii/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 in ascending from left to right. Integers in each column are sorted in...
classSolution {public:boolsearchMatrix(vector<vector<int>>& matrix,inttarget) {if(matrix.size() ==0|| matrix[0].size() ==0)returnfalse; bottom=0, top = matrix[0].size() -1;for(inti =0; i < matrix.size(); ++i) {//对每一行进行二分查找boolf =binSearch(matrix[i], bottom, ...
All the integers in each column are sorted in ascending order. -109 <= target <= 109 突然发现 LeetCode 很喜欢从 LintCode 上盗题,这是逼我去刷 LintCode 的节奏么?! 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,不满足每行的首个元素比前一行的最后一个元素大;不过对于每一行还是有序的,所以,我们可以利用分治算法,对每一行利用二分查找; Java class Solution { public boolean searchMatrix(int[][] matrix, int target) { ...
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 < ...
center > target. In this case, we discard zone 4. center == target. return true. Time Complexity: T(nxn) = 3T(n/2 x n/2) => O(3^logN) Space Complexity: O(logN) N = n^n Solution1 Code: classSolution1{publicbooleansearchMatrix(int[][]matrix,inttarget){if(matrix==null||matr...
Write an efficient algorithm that searches for a value in anmxnmatrix. 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. ...
This matrix has the following properties: Integers in each row are sorted in ascending from left to right. Integers in each c...LeetCode 240. Search a 2D Matrix II Search a 2D Matrix II 这个solution复杂度为O(row + column) 主要思想是从右上角开始搜索,行列元素值递增,所以当前元素小于...
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; ...