LeetCode 74. Search a 2D Matrix

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.

if observe carefully, it is actually using binary search..... row1, row2, row3 increasing sequentially.

#include <vector>
#include <iostream>
using namespace std;

int findMatrix(vector< vector<int> >& matrix, int pos) {
    int cols = matrix[0].size();
    int row = pos / cols;
    int col = pos % cols;
    return matrix[row][col];
}


bool searchMatrix(vector< vector<int> >& matrix, int target) {
    if(matrix.size() == 0) return false;
    if(matrix[0].size() == 0) return false;
    int left = 0;
    int right = matrix.size() * matrix[0].size() - 1;
    while(left <= right) {
        int mid = left + (right - left) / 2;
        int value = findMatrix(matrix, mid);
        if(value == target) return true;
        else if(value < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return false;
}  // time complexity O(lg(mn))

int main(void) {
    vector< vector<int> > matrix { {0, 1, 2},{3, 4, 5}, {6, 7, 8} };  // supported by -std=c++11
    bool findTarget = searchMatrix(matrix, 10);
    cout << findTarget << endl;
}