74. Search a 2D Matrix && 240. Search a 2D Matrix II

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.

For example,

Consider the following matrix:

[
[1,   3,  5,  7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]

Given target =

3

, return

true

.

Array Binary Search

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(M) Search a 2D Matrix II

public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int row_num = matrix.length;
int col_num = matrix[0].length;
//Treat it as a one-dimension array.
int begin = 0, end = row_num * col_num - 1;
while(begin <= end){
int mid = (begin + end) / 2;
int mid_value = matrix[mid/col_num][mid%col_num];

if( mid_value == target){
return true;
}else if(mid_value < target){
begin = mid+1;//Otherwise infinite loop.
}else{
end = mid-1;
}
}
return false;
}
}

240. Search a 2D Matrix II

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 in ascending from left to right.

Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[
[1,   4,  7, 11, 15],
[2,   5,  8, 12, 19],
[3,   6,  9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]

Given target =

5

, return

true

.

Given target =

20

, return

false

.

O(M+N) solution:

public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix == null || matrix.length < 1 || matrix[0].length <1) {
return false;
}
int col = matrix[0].length-1;
int row = 0;
while(col >= 0 && row <= matrix.length-1) {
if(target == matrix[row][col]) {
return true;
} else if(target < matrix[row][col]) {
col--;
} else if(target > matrix[row][col]) {
row++;
}
}
return false;
}
}

2D Binary Search. Divide and Conquer (log4/3(M*N)).

public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
return searchMatrix(matrix, target, 0, matrix.length-1, 0, matrix[0].length-1);
}

//[rFrom, rTo], [cFrom, cTo]
private boolean searchMatrix(int[][] matrix, int target, int rFrom, int rTo, int cFrom, int cTo) {
if(rFrom > rTo || cFrom > cTo)
return false;
if(rFrom == rTo && cFrom == cTo)
{
if(matrix[rFrom][cFrom] == target)
return true;
return false;
}

int rMid = (rFrom+rTo)/2;
int cMid = (cFrom+cTo)/2;
if(matrix[rMid][cMid] == target)
return true;

if(matrix[rMid][cMid] < target)
return
searchMatrix(matrix, target, rFrom, rTo, cMid+1, cTo) ||   // right side region
searchMatrix(matrix, target, rMid+1, rTo, cFrom, cMid);   //lower left region
return
searchMatrix(matrix, target, rFrom, rMid-1, cFrom, cTo) ||  //up side region
searchMatrix(matrix, target, rMid, rTo, cFrom, cMid-1);   //lower left region

}
}

M*log2(N): Do the 1D binary search for each row.