题目链接：Search a 2D Matrix
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.
这道题很简单，为此专门写篇博客其实算博客凑数了。给你一个每一行每一列都是增序，且每一行第一个数都大于上一行末尾数的矩阵，让你判断某个数在这个矩阵中是否存在。
假设矩阵是mn，扫一遍的时间复杂度就是O(mn)，题目中给出的这么特殊的矩阵，时间复杂度可以降到O(m+n)，具体代码如下，写的比较挫。
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix.length == 0)
return false;
int x = matrix.length;
int y = matrix[0].length;
int i = 0, j = 0;
while (i < x && matrix[i][j] <= target) {
i++;
}
if (i > 0) i--;
while (j < y) {
if (j < y && matrix[i][j] == target)
return true;
j++;
}
return false;
}
}
