LeetCode OJ：Search a 2D Matrix（二维数组查找）

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.

二分法的简单变形，把整个数组当成一个长的大数组就可以了，代码如下：

 class Solution {
 public:
     bool searchMatrix(vector<vector<int>>& matrix, int target) {
         int szHor = matrix.size();
         if(!szHor) return false;
         int szVer = matrix[].size();
         if(!szVer) return false;
         int totalSize = szHor * szVer;
         return bs(matrix, , totalSize - , target, szVer);

     }

     bool bs(vector<vector<int>> & matrix, int beg, int end, int target, int colCount)
     {
         if(beg > end)
             return false;
         int mid = beg + (end - beg)/;
         int val = matrix[mid/colCount][mid%colCount];
         if(val == target) return true;
         else if(val < target)
             return bs(matrix, mid + , end, target, colCount);
         else
             return bs(matrix, beg, mid - , target, colCount);
     }

 };