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;
}
};