[LintCode] 38. Search a 2D Matrix_II

public class Solution {
    /**
     * O(m + n) time complexity
     */
     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 m = matrix.length;
        int n = matrix[0].length;
        //Every time observe the point in the upper right corner
        int cur_i = 0;
        int cur_j = n - 1;
        int sum = 0;
        while(cur_i<m && cur_j >= 0){
            if(matrix[cur_i][cur_j] == target){
                sum ++;
                cur_j --; //This column can no longer satisfy the condition
            }else if(matrix[cur_i][cur_j] > target){
                cur_j --; //This column can no longer satisfy the condition
            }else{  
                cur_i ++;//The point is smaller than the target point, indicating that the line cannot meet the conditions
            }
        }
        return sum;
    }
}