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]…

leetcode_search_a_2d_matrix.cpp

bool big(vector<vector<int> > &matrix, int leftr, int leftc, int rightr, int rightc) {
    return (leftr > rightr) || (leftr == rightr && leftc > rightc);
}

void getmid(vector<vector<int> > &matrix, int leftr, int leftc, int rightr, int rightc, int& midr, int& midc) {
    int n = matrix.size();
    int m = matrix[0].size();
    int len = (rightr * m + rightc - leftr * m - leftc) / 2;
    int pos = leftr * m + leftc + len;
    midr = pos / m;
    midc = pos % m;
}

void incr(vector<vector<int> > &matrix, int midr, int midc, int &leftr, int &leftc) {
    int n = matrix.size();
    int m = matrix[0].size();
    leftr = midr;
    leftc = midc + 1;
    if (leftc == m) {
        leftr++;
        leftc = 0;
    } 
}

void desc(vector<vector<int> > &matrix, int midr, int midc, int &rightr, int &rightc) {
    int n = matrix.size();
    int m = matrix[0].size();
    rightr = midr;
    rightc = midc - 1;
    if (rightc == -1) {
        rightr--;
        rightc = m - 1;
    }
}

bool valid(vector<vector<int> > &matrix, int r, int c) {
    int n = matrix.size();
    int m = matrix[0].size();
    return r >= 0 && r < n && c >= 0 && c < m;
}

bool bsearch(vector<vector<int> > &matrix, int leftr, int leftc, int rightr, int rightc, int target) {
    if (!valid(matrix, leftr, leftc) || !valid(matrix, rightr, rightc) || big(matrix, leftr, leftc, rightr, rightc))
        return false;
    int midr, midc;
    getmid(matrix, leftr, leftc, rightr, rightc, midr, midc);
    if (matrix[midr][midc] == target)
        return true;
    else if (matrix[midr][midc] < target)
        incr(matrix, midr, midc, leftr, leftc);
    else 
        desc(matrix, midr, midc, rightr, rightc);
    return bsearch(matrix, leftr, leftc, rightr, rightc, target);
}

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