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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.
Given target = 20, return false.

Solution1：二分查找

思路： 从右上点(row=0, col=w-1)开始，以这点为中心的第一行和最后一列的整体是递增的，则可以通过二分查找去掉一行(row++)或一列(col--)，以更新的右上点继续此过程。
Time Complexity: O(m+n) Space Complexity: O(1)

Solution2：Divide and Conquer

reference: https://discuss.leetcode.com/topic/33240/java-an-easy-to-understand-divide-and-conquer-method
思路：

  zone 1      zone 2
*  *  *  * | *  *  *  *
*  *  *  * | *  *  *  *
*  *  *  * | *  *  *  *
*  *  *  * | *  *  *  *
-----------------------
*  *  *  * | *  *  *  *
*  *  *  * | *  *  *  *
*  *  *  * | *  *  *  *
*  *  *  * | *  *  *  *
  zone 3      zone 4

First, we divide the matrix into four quarters as shown below:
We then compare the element in the center of the matrix with the target. There are three possibilities:
center < target. In this case, we discard zone 1 because all elements in zone 1 are less than target.
center > target. In this case, we discard zone 4.
center == target. return true.

Time Complexity: T(nxn) = 3T(n/2 x n/2) => O(3^logN)
Space Complexity: O(logN)
N = n^n

Solution1 Code：

class Solution1 {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length < 1 || matrix[0].length < 1) {
            return false;
        }
        int row = 0;
        int col = matrix[0].length - 1;
        
        while(col >= 0 && row <= matrix.length - 1) {
            if(target == matrix[row][col]) {
                return true;
            } else if(target < matrix[row][col]) {
                col--;
            } else if(target > matrix[row][col]) {
                row++;
            }
        }
        return false;
    }
}

Solution2 Code：

class Solution2 {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length;
        if(m<1) return false;
        int n = matrix[0].length;

        return searchMatrix(matrix, new int[]{0,0}, new int[]{m-1, n-1}, target);
    }

    private boolean searchMatrix(int[][] matrix, int[] upperLeft, int[] lowerRight, int target) {
        if(upperLeft[0] > lowerRight[0] || upperLeft[1] > lowerRight[1]
                || lowerRight[0] >= matrix.length || lowerRight[1] >= matrix[0].length) 
            return false;
        if(lowerRight[0] - upperLeft[0]==0 && lowerRight[1] - upperLeft[1]==0)
            return matrix[upperLeft[0]][upperLeft[1]] == target;
        int rowMid = (upperLeft[0] + lowerRight[0]) >> 1;
        int colMid = (upperLeft[1] + lowerRight[1]) >> 1;
        int diff = matrix[rowMid][colMid] - target;
        if(diff > 0) {
            return searchMatrix(matrix, upperLeft, new int[]{rowMid, colMid}, target)
                    || searchMatrix(matrix, new int[]{upperLeft[0],colMid + 1}, new int[]{rowMid, lowerRight[1]}, target)
                    || searchMatrix(matrix, new int[]{rowMid + 1,upperLeft[1]}, new int[]{lowerRight[0], colMid}, target);
        }
        else if(diff < 0) {
            return searchMatrix(matrix, new int[]{upperLeft[0], colMid + 1}, new int[]{rowMid, lowerRight[1]}, target)
                    || searchMatrix(matrix, new int[]{rowMid + 1, upperLeft[1]}, new int[]{lowerRight[0], colMid}, target)
                    || searchMatrix(matrix, new int[]{rowMid + 1, colMid + 1}, lowerRight, target);
        }
        else return true;
    }
}