dichotomy

public class Solution {
	//The idea this time is to do a binary search on each line.
    public boolean Find(int target, int [][] array) { 
         int rows=array.length;
        int columns=array[0].length-1;
        //It's classic to be familiar with binary lookup code.
        for(int i=0;i<rows;i++){
            int low=0,high=columns;
            //Here is less than or equal to!! If the condition of self-judgment is wrongly written, it can also be carried out.
            while(low<=high){
                int mid=(low+high)/2;
                if(target<array[i][mid])
                   //That's what you remember, because the target is on the left, so high is moving to the left, so it's minus one.
                    high=mid-1;
                else if(target>array[i][mid])
                    low=mid+1;
                else 
                    return true;
                    
                
            }    
                }
        return false;
        
}
                }

Using the law of increasing from top to bottom and from left to right of two-dimensional arrays

Start at the lower left or upper right corner

public class Solution {
    public boolean Find(int [][] array,int target) {
        int row=0;
        int col=array[0].length-1;
        while(row<=array.length-1&&col>=0){
            if(target==array[row][col])
                return true;
            else if(target>array[row][col])
                row++;
            else
                col--;
        }
        return false;
 
    }
}

Attention points

Two-dimensional array

• Two-dimensional arrays in java are made up of many rows!! Two-dimensional arrays. length is rows.

dichotomy

• In dichotomy, +1, -1 are all for mid.
• You find your target (the target value you want to find), on the left of mid, let mid-1
• You find your target (the target value you want to find), on the right side of mid, let mid+1
• The condition is while (low <= high)