Leetcode 74 and 240. Search a 2D matrix I and II

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.

第一题的条件比第二题还强，也就是本行的第一个元素比上一行的最后一个元素要大。所以如果第一题把2D矩阵拉平成一个1D list，这个list是一个sorted list。可以再用二分法查找。

 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         if not matrix:
 9             return False
10             
11         nums = []
12         for elem in matrix:
13             nums += elem
14         n = len(nums)
15 
16         low, high = 0, n-1 
17         while low <= high:
18             mid = (low+high)//2
19             if target < nums[mid]:
20                 high = mid -1
21             elif target > nums[mid]:
22                 low = mid + 1
23             else:
24                 return True
25         return False

 如果不把2D matrix转换成1D array，可以直接操作row 和 col。注意L17-L18的对于从1D的位置到row col的转换方法。

 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         if not matrix:
 9             return False
10             
11         m = len(matrix)
12         n = len(matrix[0])
13         l , r = 0, n*m -1
14         
15         while l <= r:
16             mid = l + (r - l)/2
17             row = mid/ n
18             col = mid% n
19             if matrix[row][col] == target:
20                 return True
21             elif matrix[row][col] < target:
22                 l = mid + 1
23             else:
24                 r = mid - 1
25         return False
26

 

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 in ascending from left to right.
  • Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

第二题不能像第一题那样把2D matrix拉成一个1D list。可以使用类似 kth smallest element in a sorted matrix 中的类似查找方法。

 1 def searchMatrix(self, matrix, target):
 2         """
 3         :type matrix: List[List[int]]
 4         :type target: int
 5         :rtype: bool
 6         """
 7         m, n = len(matrix), len(matrix[0])
 8         i, j = m - 1, 0
 9         
10         while i >= 0 and j < n:
11             if matrix[i][j] > target:
12                 i -= 1
13             elif matrix[i][j] < target:
14                 j += 1
15             else:
16                 return True
17         return False

 

posted @ 2016-12-23 05:51  lettuan  阅读(271)  评论(0)  编辑  收藏  举报