Search a 2D Matrix II

Description

Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This matrix has the following properties:

• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Example 1:

Input: matrix = [[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]], target = 5 Output: true

Example 2:

Input: matrix = [[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]], target = 20 Output: false

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= n, m <= 300
• -109 <= matrix[i][j] <= 109
• All the integers in each row are sorted in ascending order.
• All the integers in each column are sorted in ascending order.
• -109 <= target <= 109

Code

Binary Search

Time Complexity: $O(n\log m)$, Space Complexity: $O(1)$

概念： Search a 2D Matrix 、Binary Search 101

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int n = matrix.size();
        for(int i = 0; i < n; i++) {
            int l = 0, r = matrix[i].size() - 1;
 
            while(l < r) {
                int mid = l + (r - l) / 2;
                if(matrix[i][mid] >= target) r = mid;
                else l = mid + 1;
            }
 
            if(matrix[i][l] == target) return true;
        }
        return false;
    }
};

Search from the top right

Time Complexity: $O(m+n)$, Space Complexity: $O(1)$

KEY：column 和 row 都是 sorted。

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int n = matrix.size();
        int m = matrix[0].size();
        int i = 0, j = m - 1;
        while(i >= 0 && i < n && j < m && j >= 0) {
            if(matrix[i][j] < target) i++;
            else if (matrix[i][j] > target) j--;
            else return true;
        }
        return false;
    }
};

Link

• Search a 2D Matrix II