LeetCode in Java [9]: 74. Search a 2D Matrix-优快云博客

本文链接：https://blog.youkuaiyun.com/cyr429/article/details/103636926

本文介绍了一个高效的算法，用于在一个m x n的矩阵中搜索特定值。该矩阵的每一行从左到右排序，且每行的第一个数大于前一行的最后一个数。通过全程二分查找，实现了快速定位目标值。示例展示了输入矩阵和目标值，返回是否找到目标值的布尔值。

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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;
    }
}