Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n matrix, return the occurrence of it.

This matrix has the following properties:

    * Integers in each row are sorted from left to right.

    * Integers in each column are sorted from up to bottom.

    * No duplicate integers in each row or column.

Example

Consider the following matrix:

[

    [1, 3, 5, 7],

    [2, 4, 7, 8],

    [3, 5, 9, 10]

]

Given target = 3, return 2.

Challenge

O(m+n) time and O(1) extra space

对于2d的矩阵 可以从左下角或者右上角沿着对角线找， 

例如从左下角开始找（因为从上到下和从左到右都是递增的）， 如果大于target就网上走一格， 如果小于target就往下走一格

public class Solution {
    /**
     * @param matrix: A list of lists of integers
     * @param: A number you want to search in the matrix
     * @return: An integer indicate the occurrence of target in the given matrix
     */
    public int searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0){
            return 0;
        }
        if (matrix[0] == null || matrix[0].length == 0){
            return 0;
        }
        int row = matrix.length - 1;
        int column = matrix[0].length - 1;
        int m = row;
        int n = 0;
        int count = 0;
        while (m >= 0 && m <= row && n >= 0 && n <= column){
            int cur = matrix[m][n];
            if (cur == target){
                count++;
                m--;
            } else if (cur > target){
                m--;
            } else {
                n++;
            }
        }
        return count;
    }
}