[LeetCode 74] Search a 2D Matrix

Question

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`.

Stats

 Frequency 3 Difficulty 3 Adjusted Difficulty 3 Time to use ----------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Analysis

This is a binary search question.

Solution

I did not use binary, but use the easier linear search. It still passed.

Code

my code revised (2D binary search)

``````public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return false;
}
int m = matrix.length;
int n = matrix[0].length;
// find target vertically from matrix[0] to matrix[m-1]
int top = 0, bottom = m - 1;
int mid;
while (top + 1 < bottom) {
mid = top + (bottom - top) / 2;
if (matrix[mid][0] < target) {
top = mid;
}
else {
bottom = mid;
}
}
// locate the row number
int row = -1;
if (matrix[top][0] <= target && target <= matrix[top][n-1]) {
row = top;
}
else if (matrix[bottom][0] <= target && target <= matrix[bottom][n-1]) {
row = bottom;
}
else {
return false;
}
// now find target from matrix[row]
int left = 0, right = n - 1;
while (left + 1 < right) {
mid = left + (right - left) / 2;
if (matrix[row][mid] < target) {
left = mid;
}
else {
right = mid;
}
}
return (matrix[row][left] == target || matrix[row][right] == target);
}
``````

A good binary search code here (1D binary search)

``````public boolean searchMatrix(int[][] matrix, int target) {
if(matrix==null || matrix.length==0 || matrix[0].length==0)
return false;

int m = matrix.length;
int n = matrix[0].length;
int start = 0;
int end = m*n-1;

while(start<=end){
int mid=(start+end)/2;
int midX=mid/n;
int midY=mid%n;

if(matrix[midX][midY]==target)
return true;
if(matrix[midX][midY]<target){
start=mid+1;
}else{
end=mid-1;
}
}
return false;
}
``````