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[0].size();
        if(!szVer) return false;   
        int totalSize = szHor * szVer;
        return bs(matrix, 0, totalSize - 1, 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)/2;
        int val = matrix[mid/colCount][mid%colCount];
        if(val == target) return true;
        else if(val < target)
            return bs(matrix, mid + 1, end, target, colCount);
        else 
            return bs(matrix, beg, mid - 1, target, colCount);
    }


};