文章标题

240/Search a 2D Matrix II
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
1）Integers in each row are sorted in ascending from left to right.
2）Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following matrix:
[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
Given target = 5, return true.
Given target = 20, return false.

题意：
编写一个高效的算法，查询m*n矩阵中的某个值。这个矩阵具有以下属性:
1.每一行的整数从左到右按升序排序。
2.每一列的整数从上到下按升序排序。

解题思路：二分法
先二分查找第一列，找出目标值所在的行，再对所在行进行二分查找！

代码：

class solution{public:    bool searchMattrix(vector<vector<int>>&matrix,int target)    {        int m = matrix.size();//行数        int n = matrix[0].size();//列数        int index=binary_search_1(matrix,target,0,m-1);//对第一列进行二分查找，锁定所在行            if(index<0 || index>=m)                return false;               return binary_search_2(matrix[index],target,0,n-1);//对所在行进行二分查找      }protected:    int binary_search_1(vector<vector<int>> &matrix, int target, int start, int end)    //先锁定target所在行    {        if(start >= end)            return start;        int mid = (start + end)/2;        if(target >= matrix[mid][0] && target < matrix[mid+1][0])            return mid;        else if(target < matrix[mid][0])        {             return binary_search_1(matrix,target,start,mid-1);        }        else        {            return binary_search_1(matrix,target,mid+1,end);        }    }    bool binary_search_2(vector<int> &row, int target, int start, int end)     //在锁定行中查找target      {           if(start>=end)                 return row[start]==target;           int mid=(start+end)/2;            if(row[mid]==target)                   return true;              else if(row[mid]<target)                 return binary_search_2(row,target,mid+1,end);              else                 return binary_search_2(row,target,start,mid-1);     }};