【leetcode】Search a 2D Matrix

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.

 

最简单的想法，转化成一维的，再用二分法：

 

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
       
        int row=matrix.size();
        int col=matrix[0].size();
        vector<int> m(row*col);
       
        for(int i=0;i<row;i++)
        {
            for(int j=0;j<col;j++)
            {
                m[i*col+j]=matrix[i][j];
            }
        }
       
        int left=0;
        int right=m.size()-1;
        int mid;
        while(left<=right)
        {
            mid=(left+right)/2;
            if(m[mid]>target)
                right=mid-1;
            else if(m[mid]<target)
                left=mid+1;
            else
                return true;
        }
        return false;
       
    }
};

 

 

第二种思路，直接使用二分法

 

把一维数字转化为二维的坐标的方法：

第n个元素，在n/col行，n%col列

 

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
       
        int row=matrix.size();
        int col=matrix[0].size();
       
        int left=0;
        int right=row*col-1;
        int mid;
        int m;
        while(left<=right)
        {
            mid=(left+right)/2;
            m=matrix[mid/col][mid%col];
            if(m>target)
                right=mid-1;
            else if(m<target)
                left=mid+1;
            else
                return true;
        }
        return false;          
    }
};