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`

.

**Solution:**

从右上角开始找起。time complexcity是O(M+N)

bool searchMatrix(vector<vector > &matrix, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(matrix.size()==0||matrix[0].size()==0)
return false;
int col = matrix[0].size()-1;
int row = 0;
while(row<matrix.size()&&col>=0)
{
if(target==matrix[row][col])
return true;
else if(target>matrix[row][col])//go down
row++;
else//go left
col--;
}
return false;
}

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