Write an efficient algorithm that searches for a value in an m x n matrix, return the occurrence of it.

This matrix has the following properties:

* Integers in each row are sorted from left to right.

* Integers in each column are sorted from up to bottom.

* No duplicate integers in each row or column.

Example

Consider the following matrix:

[

    [1, 3, 5, 7],

    [2, 4, 7, 8],

    [3, 5, 9, 10]

]

Given target = 3, return 2.

Challenge

O(m+n) time and O(1) extra space

Solution:

 public class Solution {
/**
* @param matrix: 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 int searchMatrix(ArrayList<ArrayList<Integer>> matrix, int target) {
int m = matrix.size();
if (m==0) return 0;
int n = matrix.get(0).size();
if (n==0) return 0; return searchMatrixRecur(matrix,target,0,0,m-1,n-1);
} public int searchMatrixRecur(ArrayList<ArrayList<Integer>> matrix, int target, int x1, int y1, int x2, int y2){
if (x2<x1 || y2<y1) return 0; if (x1==x2 && y1==y2)
if (matrix.get(x1).get(y1)==target) return 1;
else return 0; int midX = (x1+x2)/2;
int midY = (y1+y2)/2;
int midVal = matrix.get(midX).get(midY);
int res = 0; if (midVal==target){
//We have to search all the four sub matrix.
res++;
res += searchMatrixRecur(matrix,target,x1,y1,midX-1,midY-1);
res += searchMatrixRecur(matrix,target,midX+1,midY+1,x2,y2);
res += searchMatrixRecur(matrix,target,(x1+x2)/2+1,y1,x2,(y1+y2)/2-1);
res += searchMatrixRecur(matrix,target,x1,(y1+y2)/2+1,(x1+x2)/2-1,y2);
} else if (midVal>target) {
int leftX = (x1+x2)/2;
int leftY = y1;
int upX = x1;
int upY = (y1+y2)/2;
if (target==matrix.get(leftX).get(leftY)) res++;
if (target==matrix.get(upX).get(upY)) res++;
if (target <= matrix.get(leftX).get(leftY) && target <=matrix.get(upX).get(upY)){
res += searchMatrixRecur(matrix,target,x1,y1,midX-1,midY-1);
} else if (target <= matrix.get(leftX).get(leftY)){
res += searchMatrixRecur(matrix,target,x1,y1,(x1+x2)/2-1,y2);
} else if (target <= matrix.get(upX).get(upY)){
res += searchMatrixRecur(matrix,target,x1,y1,x2,(y1+y2)/2-1);
} else {
res += searchMatrixRecur(matrix,target,x1,y1,x2,(y1+y2)/2-1);
res += searchMatrixRecur(matrix,target,upX,upY,(x1+x2)/2-1,y2);
}
} else {
int rightX = (x1+x2)/2;
int rightY = y2;
int lowX = x2;
int lowY = (y1+y2)/2;
if (target==matrix.get(rightX).get(rightY)) res++;
if (target==matrix.get(lowX).get(lowY)) res++;
if (target >= matrix.get(rightX).get(rightY) && target >= matrix.get(lowX).get(lowY)){
res += searchMatrixRecur(matrix,target,midX+1,midY+1,x2,y2);
} else if (target >= matrix.get(rightX).get(rightY)){
res += searchMatrixRecur(matrix,target, (x1+x2)/2+1,y1,x2,y2);
} else if (target >= matrix.get(lowX).get(lowY)){
res += searchMatrixRecur(matrix,target, x1, (y1+y2)/2+1, x2, y2);
} else {
res += searchMatrixRecur(matrix,target, (x1+x2)/2+1,y1, lowX, lowY);
res += searchMatrixRecur(matrix,target, x1, (y1+y2)/2+1, x2, y2);
} }
return res;
}
}

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