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

时间:2015-12-23 15:49:12      阅读:156      评论:0      收藏:0      [点我收藏+]

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题目来源:

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


 

题意分析:

  给一个m×n矩阵,矩阵是按照从小到大排列,也就是a[i][j]和a[m][n]如果i  > m则a[i][j]>a[m][n],如果i == m,j > n,则a[i][j]>a[m][n]。


 

题目思路:

  利用二分查找,首先确定在那个行,然后继续二分查找。


 

代码(Python):

技术分享
 1 class Solution(object):
 2     def searchMatrix(self, matrix, target):
 3         """
 4         :type matrix: List[List[int]]
 5         :type target: int
 6         :rtype: bool
 7         """
 8         m = len(matrix)
 9         if m == 0:
10             return False
11         n = len(matrix[0])
12         if n == 0:
13             return False
14         mfirst,mlast = 0,m - 1
15         while mfirst < mlast:
16             mid = (mfirst + mlast + 1) // 2
17             if matrix[mid][0] == target:
18                 return True
19             if matrix[mid][0] < target:
20                 mfirst = mid
21             else:
22                 mlast = mid - 1
23         print(mfirst)
24         nfirst,nlast = 0,n - 1
25         while nfirst <= nlast:
26             mid = (nfirst + nlast) // 2
27             if matrix[mfirst][mid] == target:
28                 return True
29             if matrix[mfirst][mid] < target:
30                 nfirst = mid + 1
31             else:
32                 nlast = mid - 1
33         return False
View Code

 


 

转载请注明出处:http://www.cnblogs.com/chruny/p/5069802.html

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

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原文地址:http://www.cnblogs.com/chruny/p/5069802.html

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