74. Search a 2D Matrix

Problem: https://leetcode.com/problems/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

Thought:

　　search last row , last column and the left matrix (I didn't consider the second property)

　　If I consider the second property, search for row then search the row would be much better.(O(log(mn)))

Code C++:

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if (matrix.size() < 1) {
return false;
}

int m = matrix.size(), n = matrix[0].size();

if (target > matrix[m - 1][n - 1]) {
return false;
} else if (target == matrix[m - 1][n - 1]) {
return true;
}

if (m == 1) {
return binary_search(matrix[0].begin(), matrix[0].end(), target);
}
if (n == 1) {
vector<int> column;
for (int i = 0; i < matrix.size(); i++) {
column.push_back(matrix[i][0]);
}
return binary_search(column.begin(), column.end(), target);
}

vector<vector<int>> corner;
vector<int> row;
vector<int> column;

for (int i = 0; i < m - 1; i++) {
vector<int> temp;
for (int j = 0; j < n - 1; j++) {
temp.push_back(matrix[i][j]);
}
corner.push_back(temp);
}

for (int i = 0; i < n - 1; i++) {
row.push_back(matrix[m - 1][i]);
}

for (int i = 0; i < m - 1; i++) {
column.push_back(matrix[i][n - 1]);
}

return binary_search(row.begin(), row.end(), target) || binary_search(column.begin(), column.end(), target) || searchMatrix(corner, target);
}
};