leetcode74 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.

 1 class Solution { //第一个算法书写简单，O(m+n),第二个算法书写稍复杂，O(log(m)+log(n))
 2 public:
 3     bool searchMatrix(vector<vector<int>>& matrix, int target) {
 4         int m=matrix.size();
 5         if(!m)
 6             return false;
 7         int n=matrix[0].size();
 8 /*        
 9         int i=0;
10         while(i<m&&matrix[i][0]<=target)
11         {
12             i++;
13         }
14         i--;
15         if(i<0)
16             return false;
17         for(int j=0;j<n;j++)
18         {
19             if(matrix[i][j]==target)
20                 return true;
21         }
22         return false;
23 */
24         
25         int a=0,b=m-1;
26         while(a<b)
27         {
28             int mid=a+(b-a)/2;
29             int cur=matrix[mid][0];
30             if(cur==target)
31                 return true;
32             if(cur>target)
33                 b=mid;
34             else
35             {
36                 if(a==mid)
37                 {
38                     if(matrix[b][0]>target)
39                         b=a;
40                     else
41                         a=b;
42                 }
43                 else
44                     a=mid;
45             }
46         }
47         int i=a;
48         a=0;b=n-1;
49         while(a<=b)
50         {
51             int mid=a+(b-a)/2;
52             int cur=matrix[i][mid];
53             if(cur==target)
54                 return true;
55             if(cur<target)
56                 a=mid+1;
57             else
58                 b=mid-1;
59         }
60         return false;
61     }
62 };

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