leetcode 【Search a 2D Matrix 】python 实现，leetcodepython,题目：Write a

leetcode 【Search a 2D Matrix 】python 实现，leetcodepython,题目：Write a

题目：

Write an efficient algorithm that searches for a value in anmxnmatrix. 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]]

Giventarget=3, returntrue.

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 True14                 else:15                     return False16             mid=(start+end)/217             if line[mid]==target :18                 return True19             elif line[mid]>target :20                 end = mid-121             else :22                 start = mid+123 24     def searchMatrix(self, matrix, target):25         if len(matrix) == 0 :26             return False27         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 = 035         end = len(matrix)-136         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)/245             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-149             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。

leetcode 【Search a 2D Matrix 】python 实现