[Leetcode 62] 74 Search a 2D Matrix

Problem:

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.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

Analysis:

Since each row of the given matrix is sorted, we can use binary search to accelarate the search.

To locate the which to use binary search, we need to find 2 rows that row1's first element is less than target and row2's first element is greater then target. Then apply binary search to row1.

A problem here is the last row will never be searched, so we need to binary search it at last.

Code:

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (matrix.size() == 0) {
          return false;  
        } else if (matrix.size() == 1) {
            return bs(matrix[0], target);
        } else {
            for (int i=1; i<matrix.size(); i++) {
                if (matrix[i][0] == target || matrix[i-1][0] == target)
                    return true;
                else if (matrix[i][0] > target && matrix[i-1][0] < target)
                    return bs(matrix[i-1], target);
            }
        
            return bs(matrix[matrix.size()-1], target);
        }
    }
    
    bool bs(vector<int> &array, int target) {
        int lo = 0, hi = array.size()-1;
        
        while (lo <= hi) {
            int mid = (lo+hi) / 2;
            
            if (array[mid] == target) {
                return true;
            } else if (array[mid] > target) {
                hi = mid - 1;
            } else {
                lo = mid + 1;
            }
        }
        
        return false;
    }
};