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 ...
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
只有matrix[l / n][l % n] == target 时, target 才存在于矩阵中。 时间复杂度:O(log(mn)) 需要二分区间 [0, m * n - 1] ,时间复杂度为 O(log(mn)) 空间复杂度:O(1) 只需要使用常数个额外变量 代码(Python3) class Solution: def searchMatrix(self, matrix: List[List[int]], target:...
最简单的想法,转化成一维的,再用二分法: 1classSolution {2public:3boolsearchMatrix(vector<vector<int> > &matrix,inttarget) {45introw=matrix.size();6intcol=matrix[0].size();7vector<int> m(row*col);89for(inti=0;i<row;i++)10{11for(intj=0;j<col;j++)12{13m[i*col+j]=matrix[i]...
classSolution {public:boolsearchMatrix(vector<vector<int> > &matrix,inttarget) {introw =matrix.size();intcol = matrix[0].size();if(row==0||col==0)returnfalse;if(matrix[0][0]>target||target>matrix[row-1][col-1])returnfalse;//加与不加都行intlo=0,hi=row*col-1;while(lo<=hi)...
class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { if(matrix.empty()) return false; int rows=matrix.size(); int cols=matrix[0].size(); if(rows<1||cols<1) return false; int row=0; int col=cols-1; ...
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){// st...
# time O(m) + O(n)# matrix[:,end]< target,则该row所有元素均小于target# matrix[:,end] > target, 则搜索该row 可能发现target# 从右上角开始搜classSolution(object):defsearchMatrix(self,matrix,target):""" :type matrix: List[List[int]] ...
Integers in each c...LeetCode 240. Search a 2D Matrix II Search a 2D Matrix II 这个solution复杂度为O(row + column) 主要思想是从右上角开始搜索,行列元素值递增,所以当前元素小于target时,target会出现在当前行下方,当当前元素大于target时,target谁出现在当前列左侧。......
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 ...