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题目来源
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:
题意分析
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
题目思路
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。
AC代码（Python）
1 __author__ = ‘YE‘ 2 3 class Solution(object): 4 def searchMatrix(self, matrix, target): 5 """ 6 :type matrix: List[List[int]] 7 :type target: int 8 :rtype: bool 9 """ 10 m = len(matrix) 11 n = len(matrix[0]) 12 low = 0 13 high = m - 1 14 15 while low <= high: 16 mid = (low + high) / 2 17 if matrix[mid][0] == target: 18 return True 19 elif matrix[mid][0] > target: 20 high = mid - 1 21 else: 22 low = mid + 1 23 row = low - 1 24 if row < 0: 25 row = 0 26 27 low = 0 28 high = n - 1 29 30 while low <= high: 31 mid = (low + high) / 2 32 if matrix[row][mid] == target: 33 return True 34 elif matrix[row][mid] > target: 35 high = mid - 1 36 else: 37 low = mid + 1 38 return False 39 40 matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,50]] 41 target = 16 42 print(Solution().searchMatrix(matrix,target))
[LeetCode]题解（python）：074-Search a 2D Matrix
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原文地址：http://www.cnblogs.com/loadofleaf/p/5097431.html
