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.
Solution:
class Solution {
public:
bool searchMatrix(vector<vector<int> > &matrix, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
bool sol;
int col = matrix.size();
int row = matrix[0].size();
if (matrix[0][0] > target || matrix[col - 1][row - 1] < target) {
return false;
}
int col_start = 0, col_end = col - 1, col_index;
while (col_start <= col_end) {
col_index = (col_start + col_end) / 2;
if (matrix[col_index][0] == target) {
return true;
} else if (col_index != 0 && matrix[col_index][0] > target) {
if (matrix[col_index - 1][0] == target) {
return true;
}
else if (matrix[col_index - 1][0] < target) {
return search(matrix[col_index - 1], row, target);
} else {
col_end = col_index - 1;
}
} else if ((col_index != col - 1) && matrix[col_index][0] < target ) {
if (matrix[col_index + 1][0] == target) {
return true;
}
else if (matrix[col_index + 1][0] > target) {
return search(matrix[col_index], row, target);
} else {
col_start = col_index + 1;
}
} else {
// When only one row is there, then both above else-if fails.
return search(matrix[col_index], row, target);
}
}
return false;
}
bool search(vector<int> &row_mat, int row, int target) {
int beg = 0, end = row - 1, index;
//return false;
if (row_mat[row - 1] < target)
return false;
while (beg <= end) {
index = (beg + end)/2;
if (row_mat[index] == target) {
return true;
} else if (row_mat[index] > target) {
end = index - 1;
} else {
beg = index + 1;
}
}
return false;
}
};
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