LeetCode in Java [9]: 74. 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.

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

全程二分查找就行

Success

Runtime: 0 ms, faster than 100.00% of Java online submissions for Search a 2D Matrix.

Memory Usage: 41.2 MB, less than 6.06% of Java online submissions for Search a 2D Matrix.

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int cellnum=matrix.length;
        if(cellnum==0){return false;}
        int num=matrix[0].length;
        if(num==0){return false;}
        int l=0;
        int r=cellnum-1;
        while(l<=r){
            int m=(l+r)/2;
            if(matrix[m][0]<=target&&matrix[m][num-1]>=target){return bsearch(matrix[m],target,0,num-1);}
            if(matrix[m][0]>target){
                r=m-1;
                continue;
            }
            if(matrix[m][num-1]<target){
                l=m+1;
                continue;
            }
        }
        return false;
        
    }
    public boolean bsearch(int[] nums,int target,int l,int r){
        //System.out.println("enter bsearch");
        int m=(l+r)/2;
        if(nums[m]==target){return true;}
        if(l>=r){return false;}
        if(nums[m]>target){
            return bsearch(nums,target,l,m-1);
        }
        if(nums[m]<target){
            return bsearch(nums,target,l+1,r);
        }
        return false;
    }
}