42 lines
1.1 KiB
Python
42 lines
1.1 KiB
Python
|
class Solution:
|
||
|
|
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
|
||
|
|
'''
|
||
|
|
Two binary searches, first across rows and then
|
||
|
|
across columns to quickly find the target
|
||
|
|
'''
|
||
|
|
|
||
|
|
mid = -1
|
||
|
|
start, end = 0, len(matrix) - 1
|
||
|
|
|
||
|
|
# Figure out which row the number belongs to
|
||
|
|
while start <= end:
|
||
|
|
mid = (start + end) // 2
|
||
|
|
|
||
|
|
if target > matrix[mid][-1]:
|
||
|
|
start = mid + 1
|
||
|
|
elif target < matrix[mid][0]:
|
||
|
|
end = mid - 1
|
||
|
|
else:
|
||
|
|
break
|
||
|
|
|
||
|
|
# If we never managed to find a row, that means number
|
||
|
|
# doesn't exist
|
||
|
|
if not start <= end:
|
||
|
|
return False
|
||
|
|
|
||
|
|
# The row to search
|
||
|
|
row = matrix[mid]
|
||
|
|
start, end = 0, len(row) - 1
|
||
|
|
|
||
|
|
while start <= end:
|
||
|
|
mid = (start + end) // 2
|
||
|
|
|
||
|
|
if target == row[mid]:
|
||
|
|
return True
|
||
|
|
elif target > row[mid]:
|
||
|
|
start = mid + 1
|
||
|
|
else:
|
||
|
|
end = mid - 1
|
||
|
|
|
||
|
|
return False
|
||
|
|
|