Leetcode 二分查找 Search a 2D Matrix

Search a 2D Matrix

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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

复杂度：

class Solution {
public:
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;
}
};