来自专栏 · LeetCode 每日一题 题意 给定一个 m * n 的整数矩阵 matrix ,每一行从左到右升序,每一行的第一个数字比上一行的最后一个数字大,判断 target 是否在这个矩阵中? 数据限制 m == matrix.length n == matrix[i].length 1 <= m, n <= 100 -(10 ^ 4) <= matrix[i][j], target <...
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 ...
方法一: 1classSolution {2public:3boolsearchMatrix(vector<vector<int> > &matrix,inttarget)4{5introw =matrix.size();6intcol = matrix[0].size();7intsubRow =0;8if(row ==0|| col ==0)returnfalse;910//寻找行11if(matrix[row -1][0] <= target)//最后一行,特殊处理12subRow = row -...
1、二分法查找代码如下: 1publicclassSolution {2publicbooleansearchMatrix(int[][] matrix,inttarget) {3if(matrix==null|| matrix.length==0 || matrix[0].length==0)4returnfalse;56intm =matrix.length;7intn = matrix[0].length;89intstart = 0;10intend = m*n-1;1112while(start<=end){13int...
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) ...
matrix = [] for row in in_line[1].strip()[1: -4].split(']')[:-1]: matrix.append([int(n) for n in row.split('[')[1].split(',')]) k = int(in_line[2]) # print(matrix, k) print(obj.kthSmallest(matrix, k)) ...
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. ...
74. Search a 2D Matrix 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 integer of the previous row....
if matrix[i][mid] > target: r = mid 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 ...
【leetcode】240. 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 row are sorted in ascending from left to right. Integers in each c......