本文为senlie原创,转载请保留此地址:http://blog.csdn.net/zhengsenlie
Search a 2D Matrix
Total Accepted: 10871 Total
Submissions: 35629
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 row.
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3
,
return true
.
题意:在一个二维矩阵中找到给定的值。矩阵从上到下从左到右有序
思路:二维空间的二分查找
先在一维里找中间位置,再将该位置转为二维空间里的下标
注:下标比较难弄,得注意点
复杂度: 时间O(log n),空间O(1)
相关题目:
bool searchMatrix(const vector<vector<int> > &matrix, int target){ if (matrix.empty()) return false; int m = matrix.size(), n = matrix.front().size(); int begin = 0, end = m * n, middle, row, col; while(begin < end){ middle = begin + (end - begin) / 2; int row = middle / n; int col = middle % n; if(matrix[row][col] == target) return true; else if(matrix[row][col] < target) begin = middle + 1; else end = middle; } return false; }
Leetcode 二分查找 Search a 2D Matrix,布布扣,bubuko.com
时间: 2024-12-15 01:53:07