bool searchMatrix(vector<vector<int> > &matrix, int target) { if(matrix.size() == 0) return false; int m = matrix.size(); int n = matrix[0].size(); if(target < matrix[0][0] || target > matrix[m-1][n-1]) //目标元素不在矩阵中 return false; int low = 0, high = m ...
Write an efficient algorithm that searches for a value in anmxnmatrix. 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 row. For example, Consider the following matrix:...
[LeetCode]28. Search a 2D Matrix矩阵查找 Write an efficient algorithm that searches for a value in anmxnmatrix. 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 row....
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 row. For example, Consider the following matrix: [ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ] target =3, returntrue. 思路: 先搜索...
matrix = [ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ] target = 13 输出: false 这道题我采用了两种解法。一种是双层二分(二分嵌套),一种是单个二分(需要对数组的索引方式进行转换)。 双层二分的思路比较简单。先将matrix列表里的子列表当成”一个元素“,并利用每个子列表首尾...
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. ...
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...
Can you solve this real interview question? Search a 2D Matrix - You are given an m x n integer matrix matrix with the following two properties: * Each row is sorted in non-decreasing order. * The first integer of each row is greater than the last int
在LeetCode 74中,如何优化搜索二维矩阵的时间复杂度? 【原题】 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 in...
inthi=a.length()-1; while(lo<hi): intmi=lo+(hi-lo)>>1; if(a[mi]>target) lo=mi+1; else hi=mi; if(lo!=target) return-1; else returnlo; 3. 给定一个有序(非降序)数组A,可含有重复元素,求最大的i使得A[i]等于target,不存在则返回-1 ...