Leetcode 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

.

Solution:

Search from the left-right corner. If the value is greater than target, then move the pointer right to the next column; if the value is smaller than target, move the pointer up to the previous row.

Time complexity: O(m+n)

Space complexity: O(1)

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