3.二维数组中的查找
给一个有规律的整数矩阵,每行元素从左到右递增,第列元素从上到下递增。
要求完成一个查找函数,输入为满足以上条件的一个二维数组以及一个整数,判断数组中是否包含此整数。
如:
[tex] \begin{bmatrix} 1 & 2 & 8 & 9 \\ 2 & 4& 9& 12\\ 4 & 7 & 10&13 \\ 6 & 8 & 11& 15\end{bmatrix}[/tex]
查找7应当返回true,查找5应当返回false.
因为 只要求找到,不要求统计出现次数,而且此数组有规律,不用遍历用divide and conquer来做
我的习惯是用STL,但对于这个简单例子,用数组显然更高效,书中用的指针,看起来更高端了。
书中的解释很清楚。用指针访问数组元素应该是这样:
假如给定一个N-by-M的二维数组A,那么*A 指向第一个元素即A[0][0], 因为所有元素都先从行再到列地按顺序存储在内存中,所以第i行第j列的元素A[i][j]在内存中的位置是A[i*M+j]
书中从右上到左下查找,我稍微改了下,从左下到右上查找:
bool find(int *matrix, int rows, int cols, int number){ if(matrix != NULL && rows >0 && cols > 0){ // search from lower left to the upper right // start from bottom row int current_row = rows - 1; // start from first column int current_col = 0; while(current_row >=0 && current_col < cols){ if( matrix[current_row * cols + current_col] == number ){ cout<<"The number "<<number<<" is found "; cout<<"at row ("<<current_row+1<<"), column ("<<current_col+1<<")"<<endl; return true; }else{ if(matrix[current_row * cols + current_col] > number){ // go up and search above element in the same column current_row--; }else{ // go right and search element in the same row current_col++; } } } }else{ cout<<"invalid matrix (NULL pointer passed) or out of range "<<endl; return false; } cout<<number<<" is not found"<<endl; return false; }
代码:
git clone https://github.com/zhutiti/Hehaitao.git
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