"Jianzhi offer 04" two-dimensional array search

《 The finger of the sword offer 04》 Two dimensional array search

【LeetCode link 】

Two dimensional array search

【 subject 】

In a n * m In a two-dimensional array , Each row is sorted in ascending order from left to right , Each column is sorted in ascending order from top to bottom . Please complete an efficient function , Enter such a two-dimensional array and an integer , Determine whether the array contains the integer .

Example :

[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]

Given target=5, return true

Given target=20, return false

Limit ：

0 <= n <= 1000
0 <= m <= 1000

【 Ideas 】🧐

The first thought is to traverse the array , Find and return true, Go back if you can't find it false, This time complexity is O(N*M), Is there a better way ？

From the question , Each line is incremented , Each column is also incremented , So we will have the idea of binary search .

How to use the idea of bisection in two-dimensional arrays ？

We know , The core of bisection in a one bit array is to find the middle subscript mid, In a two-dimensional array , This mid Is the subscript of the last element of the previous line in every two lines .

Why? ？

Because whether we search on this line or the next line depends on target And the size of the subscript for the element value , if target Less than this element , Then we are looking for ; if target Greater than this element , Then let's go to the following line to find .

【 step 】

Pay attention here matrix.size() It's the number of lines ,matrix[0].size() It's the number of columns .

【 Code 】

class Solution {
    
public:
    bool findNumberIn2DArray(vector<vector<int>>& matrix, int target) {
    
        if (matrix.empty() == true)// If the matrix is empty , Then there must be no target
		    return false;
	    int i = 0;
	    int j = matrix[0].size()-1;
	    while (i<matrix.size() && j >= 0)
	    {
    
	    	if (matrix[i][j] == target){
    
		    	return true;
		    }
		    if (matrix[i][j]>target){
    
		    	j--;
		    }
		    else{
    
		    	i++;
		    }
	    }
	    return false;
    }
};

【 Time complexity 】

The worst case of this algorithm is to go to the lower left , So the time complexity is O(N+M).