LeetCode 74. 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

.

在一个有序的二维数组里面查找一个值，思路是先对行二分查找，再对这行二分查找，不过我的代码只二分查找了行，遍历了这一行也能AC。。

public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix.length==0||matrix[0].length==0)return false;
int row = matrix.length-1;
int col = matrix[0].length-1;
int l = 0;
int r = row;
int m = (l+r)/2;
int c = 0;
while(l<=r){
if(matrix[m][0]>target&&m!=0&&matrix[m-1][0]<=target){
c = m-1;
break;
}
if(matrix[m][col]<target&&m!=row&&matrix[m+1][col]>=target){
c = m+1;
break;
}
if(matrix[m][0]>target){
r = m-1;
m = (l+r)/2;
}
else{
l = m+1;
m = (l+r)/2;
}
}
for(int i=0;i<=col;i++){
if(matrix[c][i]==target)return true;
}
return false;
}
}