Search a 2D Matrix (Easy，二分搜索)

题目描述：

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.

举例：

Consider the following matrix:

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

Given target = 3, return true.

时间复杂度：

O(log(n) + log(m))

Code：

 1 bool searchMatrix(vector<vector<int> > &matrix, int target) {
 2         if(matrix.empty()) return false;  //矩阵为空时
 3         long long n = matrix[0].size(); //列
 4         long long m = matrix.size();    //行
 5         long long first = 0;       //0行，0列
 6         long long last = m-1;    //m-1行，0列
 7         long long mid;
 8         while(last - first >= 0){   //第1次二分，找出元素所在的行
 9             mid = first + ((last - first + 1)>>1);
10             if(matrix[mid][0] == target) return true;
11             else if(matrix[mid][0] < target){
12                 first = mid + 1;
13             }else if(matrix[mid][0] > target){
14                 last = mid - 1;
15             }
16         }
17         long long mid2;
18         if(matrix[mid][0] < target){  //第2次二分,找出元素的具体位置
19             first = 1;
20             last = n -1;
21             while(first <= last){
22                 mid2 = first + ((last - first + 1)>>1);
23                 if(matrix[mid][mid2] == target) return true;
24                 else if(matrix[mid][mid2] > target) last = mid2 - 1;
25                 else first = mid2 + 1;
26             }
27         }else if(mid >= 1 && matrix[mid][0] > target){
28             last = n - 1;
29             first = 0;
30             while(first <= last){
31                 mid2 = first + ((last - first + 1)>>1);
32                 if(matrix[mid-1][mid2] == target) return true;
33                 else if(matrix[mid-1][mid2] > target) last = mid2 - 1;
34                 else first = mid2 + 1;
35             }
36         }else
37             return false;
38
39         return false;
40     }