leetcode 【Search a 2D Matrix 】python 实现

题目：

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.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

代码：oj测试通过 Runtime: 75 ms

 1 class Solution:
 2     # @param matrix, a list of lists of integers
 3     # @param target, an integer
 4     # @return a boolean
 5     def searchInline(self, line, target):
 6         start = 0
 7         end = len(line)-1
 8         while start <= end :
 9             if start == end :
10                 return [False,True][line[start]==target]
11             if start+1 == end :
12                 if line[start]==target or line[end]==target :
13                     return True
14                 else:
15                     return False
16             mid=(start+end)/2
17             if line[mid]==target :
18                 return True
19             elif line[mid]>target :
20                 end = mid-1
21             else :
22                 start = mid+1
23 
24     def searchMatrix(self, matrix, target):
25         if len(matrix) == 0 :
26             return False
27         
28         if len(matrix) == 1 :
29             return self.searchInline(matrix[0], target)
30             
31         if len(matrix) == 2 :
32             return self.searchInline([matrix[1],matrix[0]][matrix[1][0]>target], target)\
33             
34         start = 0
35         end = len(matrix)-1
36         while start <= end :
37             if start == end:
38                 return self.searchInline(matrix[start],target)
39             if start+1 == end:
40                 if matrix[start][0] <= target and matrix[end][0] > target:
41                     return self.searchInline(matrix[start],target)
42                 if matrix[end][0] < target :
43                     return self.searchInline(matrix[end],target)
44             mid = (start+end+1)/2
45             if matrix[mid][0] <= target and matrix[mid+1][0] > target:
46                 return self.searchInline(matrix[mid],target)
47             elif matrix[mid][0] > target :
48                 end = mid-1
49             else :
50                 start = mid+1