Write an efficient algorithm that searches for a value in an m x n 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: AI检测代码解析 [[1,3,5,7],[10,11...
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-240 Search a 2D Matrix II 题目描述 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. ...
建议和leetcode 378. Kth Smallest Element in a Sorted Matrix 和 leetcode 668. Kth Smallest Number in Multiplication Table 有序矩阵搜索 代码如下: /* * 右上角搜索 * */ class Solution { public boolean searchMatrix(int[][] matrix, int target) { if(matrix==null || matrix.length<=0) retur...
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...
153. Find Minimum in Rotated Sorted Array https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/ classSolution {public:intfindMin(vector<int>&nums) {if(nums.size() ==0)return-1;inti =0, j = nums.size()-1;while(i +1<j) {intm = i + (j - i) /2;if...
0378-Kth-Smallest-Element-in-a-Sorted-Matrix 0380-Insert-Delete-GetRandom-O(1) 0381-Insert-Delete-GetRandom-O(1)-Duplicates-allowed 0382-Linked-List-Random-Node 0384-Shuffle-an-Array 0386-Lexicographical-Numbers 0387-First-Unique-Character-in-a-String 0388-Longest-Abso...
elif matrix[i][mid] < target: l = mid else: return True if target in [matrix[i][l], matrix[i][r]]: return True return False else: while l + 1 <r: mid = l + (r - l)//2 if matrix[mid][i] > target: r = mid ...