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.
Simply use binary search. Time complexity is O(logn). No extra space used.
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix.length == 0) return false;
int length = matrix.length * matrix[0].length;
int left = 0;
int right = length - 1;
while (left <= right) {
int mid = (left + right)/2;
int row = mid / matrix[0].length;
int col = mid % matrix[0].length;
if (matrix[row][col]==target){
return true;
}
else if(matrix[row][col]<target){
left = mid+1;
}
else right = mid -1;
}
return false;
}
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