[Lintcode]28. Search a 2D Matrix

28.Search a 2D Matrix

Description

Description
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 from left to right.
The first integer of each row is greater than the last integer of the previous row.
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Example
Example 1:
Input: [[5]],2
Output: false

Explanation: 
false if not included.

Example 2:
Input: [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
],3
Output: true

Explanation: 
return true if included.

Challenge
O(log(n) + log(m)) time

我的代码

class Solution:
    """
    @param matrix: matrix, a list of lists of integers
    @param target: An integer
    @return: a boolean, indicate whether matrix contains target
    """

    def searchMatrix(self, matrix, target):
        # write your code here
        if matrix == [] or matrix == [[]]:
            return False
        if matrix[0][0] > target:
            return False
        if matrix[-1][-1] < target:
            return False
        r, c = len(matrix), len(matrix[0])
        if matrix[-1][0] <= target:
            rp = r - 1
        else:
            rl = 0
            rr = r - 1
            rm = int((rl + rr) / 2)
            while rl < rm and rr > rm:
                if matrix[rm][0] == target:
                    return True
                else:
                    if matrix[rm][0] > target:
                        rr = rm
                        rm = int((rl + rr) / 2)
                    else:
                        rl = rm
                        rm = int((rl + rr) / 2)
            rp = rm
        cl = 0
        cr = c - 1
        cm = int((cl + cr) / 2)
        if matrix[rp][cl] == target or matrix[rp][cr] == target:
            return True
        while cl < cm and cm < cr:
            if matrix[rp][cm] == target:
                return True
            else:
                if matrix[rp][cm] > target:
                    cr = cm
                    cm = int((cl + cr) / 2)
                else:
                    cl = cm
                    cm = int((cl + cr) / 2)
        return False

别人的代码

class Solution:
    # @param matrix, a list of lists of integers
    # @param target, an integer
    # @return a boolean
    # 8:21
    def searchMatrix(self, matrix, target):
        if not matrix or target is None:
            return False

        rows, cols = len(matrix), len(matrix[0])
        low, high = 0, rows * cols - 1
        
        while low <= high:
            mid = (low + high) / 2
            num = matrix[mid / cols][mid % cols]

            if num == target:
                return True
            elif num < target:
                low = mid + 1
            else:
                high = mid - 1
        
        return False

思路：

我先对第一列做二分查找，再对找到的行做二分查找，但是我的特殊情况太多了，搞得代码非常不简洁非常不好看。
其实可以把它看作一个数组，则只需要一个二分查找，而且时间复杂度相同。

• 时间复杂度: O(log(m)+log(n))
• 出错：一开始的条件是cr<cl，这个条件很多情况下始终能满足，应该使用cr<rm &&cl>rm这种才能够达到预期的作用。