Lintcode Medium Title: Search a 2d Matrix II searching for two-dimensional matrices II

Topic

search for two-dimensional Matrix II

Write an efficient algorithm to search for values in the MXN matrix, returning the number of occurrences of this value.

This matrix has the following characteristics:

• The integers in each row are sorted from left to right.
• The integers of each column are sorted from top to bottom.
• There are no duplicate integers in each row or column.

Sample Example

Consider the following matrices:

[

[1, 3, 5, 7],

[2, 4, 7, 8],

[3, 5, 9, 10]

]

Give target = 3, return 2

challenges

Requires O (m+n) time complexity and O (1) Extra space

Solving

Direct traversal with a time complexity of O (MN)

 Public classSolution {/**     * @parammatrix:a List of lists of integers *@param: A number you want to search in the matrix *@return: An integer indicate the occurrence of target in the given matrix*/     Public intSearchmatrix (int[] Matrix,inttarget) {        //Write your code here        if(Matrix = =NULL)            return0; introw =matrix.length; if(Row ==0)            return0; intCol = matrix[0].length; intCount =0;  for(inti=0;i< Row; i++){             for(intj=0;j<col;j++){                if(Matrix[i][j] = =target) Count++; }        }        returncount; }}

Java Code

The array in question is ordered, and the elements of each row or column are not duplicated.

It can be found that the number of occurrences occurs between 0 and min (col,row)

There seems to be a very similar question on the offer.

By dividing the matrix, go one column at a time

Java programs

 Public classSolution {/**     * @parammatrix:a List of lists of integers *@param: A number you want to search in the matrix *@return: An integer indicate the occurrence of target in the given matrix*/     Public intSearchmatrix (int[] Matrix,inttarget) {        //Write your code here        if(Matrix = =NULL)            return0; introw =matrix.length; if(Row ==0)            return0; intCol = matrix[0].length; intCount =0; intI=0; intJ=col-1;  while(I<row && j>=0){            if(Matrix[i][j] >target) {J--; }Else if(matrix[i][j]<target) {i++; }Else{Count++; I++; J--; }        }        returncount; }    }

Java Code

Python program

classSolution:"""@param matrix:an List of lists of integers @param target:an integer you want to search in matrix @return: An integer indicates the total occurrence of target in the given matrix"""    defSearchmatrix (self, Matrix, target):#Write your code here        ifMatrix = =None:return0 Row=len (Matrix)ifrow = =0:return0 Col=Len (matrix[0]) Count=0 I=0 J= Col-1 whileI<row andj>=0:ifMATRIX[I][J] >target:j-=1elifMATRIX[I][J] <target:i+ = 1Else: Count+ = 1I+ = 1J-= 1returnCount

Python Code

Lintcode Medium Title: Search a 2d Matrix II searching for two-dimensional matrices II