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
思路：
先按行二分查找，再按列二分查找。
代码写的比较繁琐。
