解题思路：

两次二分，第一次定位行，第二次定位列。

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length, n = matrix[0].length;
        int l = 0, r = m - 1;
        
        //定位行
        int row = -1;
        while (l <= r) {
            int mid = l + (r - l) / 2;
            if (matrix[mid][0] <= target && matrix[mid][n - 1] >= target) {
                row = mid;
                break;
            } else if (target < matrix[mid][0]) {
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        if (row == -1) return false;
        
        //定位列
        l = 0;
        r = n - 1;

        while (l <= r) {
            int mid = l + (r - l) / 2;
            //已知行row之后，就按照常规的二分来做
            if (target == matrix[row][mid])
                return true;
            else if (target < matrix[row][mid])
                r = mid - 1;
            else
                l = mid + 1;
        }
        return false;
    }
}