Leetcode "Search a 2D Matrix" Python implementation

title :

Write an efficient algorithm, searches for a value in a m x n Matrix. This matrix has the following properties:

• Integers in each row is sorted from the left to the right.
• The first integer of each row was greater than the last integer of the previous row.

For example,

Consider the following matrix:

[  [1,   3,  5,  7],  [Ten, One,],  [23, 30, 34, 50]]

Given target = 3 , return true .

code : OJ Test via runtime:75 ms

1 classSolution:2     #@param Matrix, a list of lists of integers3     #@param target, an integer4     #@return A Boolean5     defsearchinline (self, line, target):6Start =07end = Len (line)-18          whileStart <=End:9             ifStart = =End:Ten                 return[False,true] [line[start]==Target] One             ifStart+1 = =End: A                 ifLine[start]==targetorline[end]==Target: -                     returnTrue -                 Else: the                     returnFalse -Mid= (start+end)/2 -             ifline[mid]==Target: -                 returnTrue +             elifLine[mid]>Target: -End = Mid-1 +             Else : AStart = Mid+1 at  -     defSearchmatrix (self, Matrix, target): -         ifLen (Matrix) = =0: -             returnFalse -          -         ifLen (Matrix) = = 1 : in             returnSelf.searchinline (matrix[0], target) -              to         ifLen (Matrix) = = 2 : +             returnSelf.searchinline ([matrix[1],matrix[0]][matrix[1][0]>target], target) -              theStart =0 *end = Len (matrix)-1 $          whileStart <=End:Panax Notoginseng             ifStart = =End: -                 returnself.searchinline (matrix[start],target) the             ifStart+1 = =End: +                 ifMATRIX[START][0] <= Target andMatrix[end][0] >Target: A                     returnself.searchinline (matrix[start],target) the                 ifMatrix[end][0] <Target: +                     returnself.searchinline (matrix[end],target) -Mid = (start+end+1)/2 $             ifMATRIX[MID][0] <= Target andMatrix[mid+1][0] >Target: $                 returnself.searchinline (matrix[mid],target) -             elifMatrix[mid][0] >Target: -End = Mid-1 the             Else : -Start = Mid+1

Ideas :

Search by row Two, then by column two points.

Code writing is cumbersome.

Leetcode "Search a 2D Matrix" Python implementation