Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3
,
return true
.
題意:在一個二維矩陣中找到給定的值。矩陣從上到下從左到右有序
思路:二維空間的二分查找
先在一維裡找中間位置,再將該位置轉為二維空間裡的下標
復雜度:
class Solution { public: bool searchMatrix(const vector> &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; } };