Title
Note: the same problem with the master 240 according to the present topic: https: //leetcode-cn.com/problems/search-a-2d-matrix-ii/
Source: stay button (LeetCode)
Link: https: // leetcode-cn. com / problems / er-wei- shu-zu-zhong-de-cha-zhao-lcof
Solution a: violence, traversing two-dimensional array, the time complexity of m * n
class Solution:
def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
if matrix==None or len(matrix)==0:
return False
rows = len(matrix) # 矩阵行数
columns = len(matrix[0]) # 矩阵列数
for i in range(rows):
for j in range(columns):
if matrix[i][j]==target:
return True
return False
Solution two: two-dimensional array because each row from left to right in order of ascending sort, each line traversing two-dimensional array, use a binary search to find methods for each row
class Solution:
def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
if matrix==None or len(matrix)==0:
return False
rows = len(matrix)
for i in range(rows):
start = 0
end = len(matrix[i])-1
while start <= end: # 对每一行用二查找方法查找
mid = start+(end-start)//2
if matrix[i][mid]>target:
end = mid-1
elif matrix[i][mid]<target:
start = mid+1
else:
return True
return False
Solution three: start from the bottom left to determine, if the lower left corner of numbers larger than the target, delete the row; on the contrary, where the delete column, gradually approaching the upper right corner
class Solution:
def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
if matrix==None and len(matrix)==0:
return False
rows = len(matrix)-1
columns = 0
while rows>=0 and columns<=len(matrix[0])-1:
if matrix[rows][columns]>target:
rows -= 1
elif matrix[rows][columns]<target:
columns += 1
else:
return True
return False
