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 ...
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...
find k-th element in the matrix:https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/ a) heap-based, starting from top-left corner element, push to heap; each time an element is poped, push its right and down elements. complexity is k*log(k) b) binary-search on t...
2.Find Peak Element View Code 能进入Line 5的while循环说明至少有3个元素。 八、 1.Remove Duplicates from Sorted Array View Code 2.Remove Duplicates from Sorted Array II code 1: [推荐] View Code 该代码是通用版代码。k表示单个元素最多能允许duplicate的个数。将k改为1即可用于上一题。 code 2: ...
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...
0023-Merge-k-Sorted-Lists 0024-Swap-Nodes-in-Pairs 0025-Reverse-Nodes-in-k-Group 0026-Remove-Duplicates-from-Sorted-Array 0027-Remove-Element 0028-Implement-strStr 0033-Search-in-Rotated-Sorted-Array/cpp-0033 CMakeLists.txt main.cpp 0034-Search-for-a-Range 0036-Va...
今天的笔记包含基于改造二分法(Modified Binary Search)类型下的7个题目,它们在leetcode上的编号和题名分别是: 704 - 二分查找 74 - 搜索二维矩阵 33 - 搜索旋转排序数组 81 - 搜索旋转排序数组 II 153 - 寻找旋转排序数组中的最小值 162 - 寻找峰值 ...
bool searchMatrix(vector<vector<int>>& matrix, int target) { if (matrix.size() == 0|| matrix[0].size() ==0) { return false; } int i = 0, j = matrix[i].size() - 1; while (i<matrix.size() && j>=0) { int element = matrix[i][j]; ...
相关LeetCode题: 704. Binary Search 题解 34. Find First and Last Position of Element in Sorted Array 题解 33. Search in Rotated Sorted Array 题解 528. Random Pick with Weight 题解 按值范围折半查找 折半查找还可以应用于非有序区间查找满足特定条件的值。该场景下所找的值在已知范围内,这时折...