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。

 class Solution(object):
     def searchMatrix(self, matrix, target):
         """
         :type matrix: List[List[int]]
         :type target: int
         :rtype: bool
         """
         if not matrix:
             return False

         nums = []
         for elem in matrix:
             nums += elem
         n = len(nums)

         low, high = 0, n-1
         while low <= high:
             mid = (low+high)//2
             if target < nums[mid]:
                 high = mid -1
             elif target > nums[mid]:
                 low = mid + 1
             else:
                 return True
         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。

 class Solution(object):
     def searchMatrix(self, matrix, target):
         """
         :type matrix: List[List[int]]
         :type target: int
         :rtype: bool
         """
         if not matrix:
             return False

         m = len(matrix)
         n = len(matrix[0])
         l , r = 0, n*m -1

         while l <= r:
             mid = l + (r - l)/2
             row = mid/ n
             col = mid% n
             if matrix[row][col] == target:
                 return True
             elif matrix[row][col] < target:
                 l = mid + 1
             else:
                 r = mid - 1
         return False
             

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。

 def searchMatrix(self, matrix, target):
         """
         :type matrix: List[List[int]]
         :type target: int
         :rtype: bool
         """
         m, n = len(matrix), len(matrix[0])
         i, j = m - 1, 0

         while i >= 0 and j < n:
             if matrix[i][j] > target:
                 i -= 1
             elif matrix[i][j] < target:
                 j += 1
             else:
                 return True
         return False