lintcode38- 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

从左下角开始向右上方向漫游。找小的往上，找大的往右。

这题和I方法上没大关联（从时间复杂度要求的类型也可以看出来），主要是根据条件的规律想到这个点子。

public class Solution {
    /*
     * @param matrix: A list of lists of integers
     * @param target: An integer you want to search in matrix
     * @return: An integer indicate the total occurrence of target in the given matrix
     */
    public int searchMatrix(int[][] matrix, int target) {
        // write your code here
        if(matrix == null || matrix.length == 0){
            return 0;
        }
        
        if(matrix[0] == null || matrix[0].length == 0){
            return 0;
        }
        
        int rows = matrix.length;
        int cols = matrix[0].length;
        
        int y = rows - 1;
        int x = 0;
        int count =0;
        
        while (y >= 0 && x < cols){
            int num = matrix[y][x];
            
            if (target == num){
                ++count;
                ++x;
                --y;
            }
            else if (target > num){
                ++x;
            }
            else {
                --y;
            }
        }
        
        return count;
    }
}