The first 74 questions: two-dimensional matrix search

One. Problem Description

Prepared by an efficient algorithm to determine the mxn matrix, the presence or absence of a target value. This matrix has the following characteristics:

An integer of from left to right in each row in ascending order.

The first integer is greater than the last row of each integer previous row.

Example 1:

Input:

matrix = [

  [1,   3,  5,  7],

  [10, 11, 16, 20],

  [23, 30, 34, 50]

]

target = 3

Output: true

Example 2:

Input:

matrix = [

  [1,   3,  5,  7],

  [10, 11, 16, 20],

  [23, 30, 34, 50]

]

target = 13

Output: false

two. Problem-solving ideas

Problem-solving ideas: the method of binary search to find

Step one: First, the first column binary search to find the number of possible targets in which row.

Step two: target row binary search to determine whether there is a target number (this question Step Two lazy, direct traversal).

 

three. Results of the

When execution: 1 ms, beat the 78.93% of all users in java submission

Memory consumption: 42.7 MB, defeated 38.63% of all users in java submission

four. Java code

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
            if(matrix==null||matrix.length<=0||matrix[0].length<=0)
        {
            return false;
        }
     int row=0;
     int first=0;
     int second=matrix.length-1;
     
     
     while(first<=second)
     {
         row=(first+second)/2;
         if((row<matrix.length-1&&(matrix[row][0]<=target&&matrix[row+1][0]>target))||((row==matrix.length-1)&&(matrix[row][0]<=target))){
           break;
         }
         else 
         {
             if(matrix[row][0]>target)
             {
                 second=row-1;
             }
             else 
             {
                 first=row+1;
             }
         }
     }
     
    
     for(int i=0;i<matrix[row].length;i++)
     {
         if(matrix[row][i]==target)
         {
             return true;
         }
     }
     return false;
    }
}