問題來源
在一個 n * m 的二維數組中,每一行都按照從左到右遞增的順序排序,每一列都按照從上到下遞增的順序排序。請完成一個函數,輸入這樣的一個二維數組和一個整數,判斷數組中是否含有該整數。
示例:
現有矩陣 matrix 如下:
[
[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]
]
給定 target = 5,返回 true。
給定 target = 20,返回 false。
限制:
0 <= n <= 1000
0 <= m <= 1000
常規思路
class Solution:
def findNumberIn2DArray(self, matrix, target: int) -> bool:
row = len(matrix)
if row == 0:
return False
if row == 1:
return target in matrix[0]
col = len(matrix[0])
for i in range(row):
for j in range(col):
if matrix[i][j] == target:
return True
return False
class Solution:
def binarySerach(self, array, target):
left, right = 0, len(array) - 1
while left <= right:
mid = int(left + (right - left) / 2)
if array[mid] == target:
return True
elif array[mid] < target:
left = mid + 1
else:
right = mid - 1
return False
def findNumberIn2DArray(self, matrix, target: int) -> bool:
row = len(matrix)
if row == 0:
return False
if row == 1:
return target in matrix[0]
col = len(matrix[0])
for i in range(row):
if self.binarySerach(matrix[i], target):
return True
return False
大佬解析
代碼
class Solution:
def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
i, j = len(matrix) - 1, 0
while i >= 0 and j < len(matrix[0]):
if matrix[i][j] > target: i -= 1
elif matrix[i][j] < target: j += 1
else: return True
return False