Count Negative Numbers in a Sorted Matrix
<p>Given a <code>m x n</code> matrix <code>grid</code> which is sorted in non-increasing order both row-wise and column-wise, return <em>the number of <strong>negative</strong> numbers in</em> <code>grid</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]
<strong>Output:</strong> 8
<strong>Explanation:</strong> There are 8 negatives number in the matrix.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[3,2],[1,0]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>-100 &lt;= grid[i][j] &lt;= 100</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you find an <code>O(n + m)</code> solution?
Array; Binary Search; Matrix
class Solution {
    public int countNegatives(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int ans = 0;
        for (int i = m - 1, j = 0; i >= 0 && j < n;) {
            if (grid[i][j] < 0) {
                ans += n - j;
                --i;
            } else {
                ++j;
            }
        }
        return ans;
    }
}