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;    }}