Problem
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.
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.
Solution
Solution 1
class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
if not matrix or not matrix[0]:
return False
j = 0
for row in matrix:
if row[j] > target:
while row[j] > target:
j -= 1
if j < 0:
return False
elif row[j] < target:
while row[j] < target:
j += 1
if j == len(matrix[0]):
j = 0
break
if row[j] == target:
return True
return False
Solution 2: Like a binary tree
Start from upper right corner.
If matrix[i][j] > target: col - 1
if matrix[i][j] < target: row + 1
class Solution:
def searchMatrix(self, matrix, target):
if not matrix or not matrix[0]:
return False
i, j = 0, len(matrix[0]) - 1
while i < len(matrix) and j >= 0 :
if matrix[i][j] == target:
return True
elif matrix[i][j] > target:
j -= 1
else:
i += 1
return False