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 ...
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 -...
class Solution: def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: # 矩阵行数 m: int = len(matrix) # 矩阵列数 n: int = len(matrix[0]) # 二分区间左边界初始化为 0 l: int = 0 # 二分区间右边界初始化为 m * n - 1 r: int = m * n - 1 # 当二分...
classSolution {public:boolsearchMatrix(vector<vector<int>>& matrix,inttarget) {if(matrix.size() ==0|| matrix[0].size() ==0)returnfalse; bottom=0, top = matrix[0].size() -1;for(inti =0; i < matrix.size(); ++i) {//对每一行进行二分查找boolf =binSearch(matrix[i], bottom, ...
LeetCode_240. Search a 2D Matrix II 题目描述: 思路:题目要求写一个高效率的算法,所以暴力遍历整个矩阵查找 的方法行不通了。 分析题目,发现这个矩阵是从左到右,从上到下递增的。可以发现矩阵右上角的元素是一个特殊的元素( ):如果 和 正好相等了,说明找到了。而当...
Solution1 Code: classSolution1{publicbooleansearchMatrix(int[][]matrix,inttarget){if(matrix==null||matrix.length<1||matrix[0].length<1){returnfalse;}introw=0;intcol=matrix[0].length-1;while(col>=0&&row<=matrix.length-1){if(target==matrix[row][col]){returntrue;}elseif(target<matrix...
publicclassSolution{/* https://leetcode.com/discuss/48852/my-concise-o-m-n-java-solution */publicbooleansearchMatrix(int[][]matrix,inttarget){if(matrix==null||matrix.length<1||matrix[0].length<1){returnfalse;}introw=0,col=matrix[0].length-1;while(row<matrix.length&&col>=0){// star...
leetcode 240. Search a 2D Matrix II 思路:从左下或者右上开始进行搜索,因为比如从左下开始搜索,若目标数大于此时的数,接下来只能向右搜,若小于,接着只能向上搜。若是从左上开始进行搜索,若目标数大于此时的数,会有两个方向可以走。 ... 240. Search a 2D Matrix II ...
Problem # 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 ...