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