【leetcode】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.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true

题解：直接把二维数组看成一个长度为m*n的数列，然后用二分法查找target，注意在这个数列中编号为k的数在二维数组中的坐标x = k/n, y=k%n;

代码如下;

 1 public class Solution {
 2     public boolean searchMatrix(int[][] matrix, int target) {
 3         if(matrix.length == 0 || matrix[0].length == 0)
 4             return false;
 5         int m = matrix.length;
 6         int n = matrix[0].length;
 7         
 8         int start = 0,end = m*n-1;
 9         while(start <= end){
10             int mid = start + (end-start)/2;
11             int x = mid/n;
12             int y = mid%n;
13             
14             if(matrix[x][y] == target)
15                 return true;
16             
17             if(matrix[x][y] < target)
18                 start = mid+1;
19             else
20                 end = mid-1;
21         }
22         return false;
23     }
24 }