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 ...
if (target < matrix[mid][0]) { right = mid; } else if (target > matrix[mid][0]) { left = mid; } else { return true; } } left = 0; right = matrix[0].length - 1; while (left <= right) { //搜索在那一列 mid = left + (right - left) / 2; if (target < matrix[...
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 <...
*@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...
如果target是和matrix[i][n-1]比较来确定target所在行的话,则上面代码注释地方恰好要使用matrix[above]了: classSolution {public:boolsearchMatrix(vector<vector<int>>& matrix,inttarget) {if(matrix.empty() || matrix[0].empty())returnfalse;if(target < matrix.front().front() || target > matrix....
https://leetcode.com/problems/search-a-2d-matrix-ii/description/ 题解一 九章算法提供的比较巧妙的解题思路,从matrix 左下角开始向右上角遍历, 初始: introwIndex=matrix.length-1;intcolIndex=0; 逻辑主体: ifcurrentItem == target, return true; ...
Consider the following matrix: [ [1, 4, 7, 11, 15], [2, 5, 8, 12, 19], [3, 6, 9, 16, 22], [10, 13, 14, 17, 24], [18, 21, 23, 26, 30] ] Giventarget=5, returntrue. Giventarget=20, returnfalse. 左上角最小右下角最大, 先让raw =0; col = n-1; 如果target...
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
大家可以看完讲义结合 LeetCode Book 二分查找练习一下 问题定义 给定一个由数字组成的有序数组 nums,并给你一个数字 target。问 nums 中是否存在 target。如果存在, 则返回其在 nums 中的索引。如果不存在,则返回 - 1。 这是二分查找中最简单的一种形式。当然二分查找也有很多的变形,这也是二分查找容易出错...
7. (Leetcode: Search Insert Position)给定一个有序(非降序)数组A,若target在数组中出现,返回位置,若不存在,返回它应该插入的位置。 intsearch(inta[]):// return index intlo=0; inthi=a.length()-1; while(lo<hi): intmi=lo+(hi-lo)/2; ...