【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[].size();
         vector<int> m(row*col);

         for(int i=;i<row;i++)
         {
             for(int j=;j<col;j++)
             {
                 m[i*col+j]=matrix[i][j];
             }
         }

         int left=;
         int right=m.size()-;
         int mid;
         while(left<=right)
         {
             mid=(left+right)/;
             if(m[mid]>target)
                 right=mid-;
             else if(m[mid]<target)
                 left=mid+;
             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[].size();

         int left=;
         int right=row*col-;
         int mid;
         int m;
         while(left<=right)
         {
             mid=(left+right)/;
             m=matrix[mid/col][mid%col];
             if(m>target)
                 right=mid-;
             else if(m<target)
                 left=mid+;
             else
                 return true;
         }
         return false;
     }
 };