74. Search a 2D Matrix

Medium (Trung bình) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

You are given an m x n integer matrix matrix with the following two properties:

  • Each row is sorted in non-decreasing order.
  • The first integer of each row is greater than the last integer of the previous row.

Given an integer target, return true if target is in matrix or false otherwise.

You must write a solution in O(log(m * n)) time complexity.

 

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

 

Constraints:

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 100
  • -104 <= matrix[i][j], target <= 104

🧠 Thuật Toán & Kỹ Thuật

Binary Search (Tìm kiếm nhị phân)Bit Manipulation (Thao tác bit)Matrix (Ma trận)
⏱️ Thời gian O(log n)
💾 Không gian O(n)

💻 Lời Giải

C++ 0074-search-a-2d-matrix.cpp 
class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int row = matrix.size(), col = matrix[0].size();
        int left = 0, right = row*col - 1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            int num = matrix[mid / col][mid % col];
            
            if (num == target) {
                return true;
            }
            else if (num < target) {
                left = mid + 1;
            }
            else {
                right = mid - 1;
            }
        }
        
        return false;
    }
};