LeetCode - 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.
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.

观察可知越往右下数越大，加上每一行最后一个数是该行最大值，因此可以根据列值减少搜索的列数。定义 i 为矩阵的行，j 为矩阵的列，初始化 i 为0，j 为最后一列。若目标值大于当前matrix [ i ][ j ] 时，说明此行不可能比目标值大，因此 i+1；若小于，则目标值必然不在此列，因此 j-1；若等于则输出。

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