[LeetCode]题解（python）：074-Search a 2D Matrix

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题目来源

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https://leetcode.com/problems/search-a-2d-matrix/

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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题意分析

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Input:  a matrix and a target to query

Output: True or False

Conditions:在矩阵中查找元素，注意矩阵是有序的

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题目思路

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先在列二分查找定位哪一列，然后再行二分查找，注意边界条件

PS:其实直接在列查找的时候，发现low-1小于0时，直接返回False就可以了，但是所贴代码可以移植到插入元素的二分排序里面，为了统一就没有改动了。

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AC代码（Python）

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 __author__ = 'YE'

 class Solution(object):
     def searchMatrix(self, matrix, target):
         """
         :type matrix: List[List[int]]
         :type target: int
         :rtype: bool
         """
         m = len(matrix)
         n = len(matrix[0])
         low = 0
         high = m - 1

         while low <= high:
             mid = (low + high) / 2
             if matrix[mid][0] == target:
                 return True
             elif matrix[mid][0] > target:
                 high = mid - 1
             else:
                 low = mid + 1
         row = low - 1
         if row < 0:
             row = 0

         low = 0
         high = n - 1

         while low <= high:
             mid = (low + high) / 2
             if matrix[row][mid] == target:
                 return True
             elif matrix[row][mid] > target:
                 high = mid - 1
             else:
                 low = mid + 1
         return False

 matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,50]]
 target = 16
 print(Solution().searchMatrix(matrix,target))