分析:将二维数组看为一个一维sorted数组,然后用bianry search。时间复杂度为O(log(m*n)),代码如下: 1classSolution {2public:3boolsearchMatrix(vector<vector<int> > &matrix,inttarget) {4intm =matrix.size();5if(m ==0)returnfalse;6intn = matrix[0].size();7if(n ==0)returnfalse;89intl =...
*@return*/publicbooleansearchMatrix(int[][] matrix,inttarget) {if(matrix ==null)returnfalse;introw =matrix.length;if(row == 0)returnfalse;if(matrix[0] ==null)returnfalse;intcolumn = matrix[0].length;if(column == 0)returnfalse;if(target < matrix[0][0] && target > matrix[row - 1...
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
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 ...
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 < ...
[LeetCode]Search a 2D Matrix Question 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 ...
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; ...
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: Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 Integers in each row are sorted from left to right. The first integer of each row is greater than the last integer...
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...