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.

解题思路

二分搜索。把矩阵看成是一维数组，建立一维数组序号与二维数组坐标之间的对应关系：
m=(pos−1)/col;
n=(pos−1)%col;
然后进行普通的二分搜索。

实现代码

// Runtime: 1 ms
public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix.length == 0) {
            return false;
        }
        int row = matrix.length;
        int col = matrix[0].length;
        int left = 1;
        int right = row * col;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            int m = (mid - 1) / col;
            int n = (mid - 1) % col;
            if (matrix[m][n] > target) {
                right = mid - 1;
            } else if (matrix[m][n] < target) {
                left = mid + 1;
            } else {
                return true;
            }
        }

        return false;
    }
}