LintCode 38. 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.

Have you met this question in a real interview?

Yes

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

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解题思路：

从左下角往右上角找，若是小于target就往右找，若是大于target就往上找。时间复杂度O(m+n)  n 为行数，m为列数。

定义count 计数。

与 Leetcode 240. Search a 2D Matrix II 类似。

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Java code:

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) {
         //check corner case
        if (matrix == null || matrix.length == 0) {
            return 0;
        }
        if (matrix[0] == null || matrix[0].length == 0) {
            return 0;
        }
        //find from bottom left to top right
        int n = matrix.length; //row
        int m = matrix[0].length; //column
        int x = n - 1;
        int y = 0;
        int count = 0;
        while (x >= 0 && y < m) {
            if (matrix[x][y] < target) {
                y++;
            } else if (matrix[x][y] > target) {
                x--;
            } else {
                count++;
                x--;
                y++;
            }
        }
        return count;
    }
}

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Reference:

1. http://www.jiuzhang.com/solutions/search-a-2d-matrix-ii/