6.2.3 python二分查找算法及LeetCode题目(3)之二维数组 —— Search a 2D Matrix
接下来是两道二维数组查找的题目，看一下二维数组中二分查找的特点，或者说应用。
74. Search a 2D Matrix
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 from left to right.
- The first integer of each row is greater than the last integer of the previous row.
Example 1:
Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output: true题目解析：
该题二维数组的特点是，按顺序拼接为一维数组后，也仍然是有序的，因此思路也很简单，先按每行第一个元素二分查找，确定目标元素在哪一行；再在该行进行二分查找。代码如下，做了许多边界条件处理，性能不错，代码有许多优化空间，不过了解思路即可。
由于前面所说，该二维数组拼为一维，也仍然是有序，所以可以把2D数组当成一维的，通过 // 和 %计算行和列，只进行一次二分查找即可。
class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
m = len(matrix)
if not m:
return False
n = len(matrix[0])
if not n:
return False
st = matrix[0][0]
ed = matrix[-1][-1]
if st > target or ed < target:
return False
lt, rt = 0, m-1
pos = lt
while lt <= rt:
mid = (lt + rt) // 2
# print(mid)
num_st = matrix[mid][0]
if num_st <= target and (mid == m-1 or target < matrix[mid+1][0]):
pos = mid
break
elif num_st > target:
rt = mid - 1
else:
lt = mid + 1
# print(pos)
lt, rt = 0, n-1
if matrix[pos][0] > target or matrix[pos][-1] < target:
return False
while lt <= rt:
mid = (lt + rt) // 2
# print(mid)
num = matrix[pos][mid]
if num == target:
return True
elif num > target:
rt = mid - 1
else:
lt = mid + 1
return False240. Search a 2D Matrix II
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.
题目解析：
此题的2D数组就是真二维了，自上而下自左到右都有序。鄙人介绍提供如下两种解题方法：
一， 仿照上题思路，通过每行第一个和最后一个元素，确定target可能会在的某几行（因此用了两个while循环）；然后在这几行中分别进行二分查找（for循环中一个while循环），整体上时间复杂度是2*O(logm)+mlog(n)，但是实际上性能还不错。
二，是一个值得思考和学习的方法，从后面的Discus部分学习来的，时间复杂度O(m+n)，从第一行最后一个元素开始查找，大于target光标左移，小于target光标下移，自己看代码思考一下吧~
class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
m = len(matrix)
if not m:
return False
n = len(matrix[0])
if not n:
return False
if matrix[0][0] > target or matrix[-1][-1] < target:
return False
lt, rt = 0, m-1
while lt <= rt:
mid = (lt + rt) // 2
m_num = matrix[mid][0]
if m_num <= target and (mid == m-1 or matrix[mid+1][0] > target):
break
elif m_num > target:
rt = mid - 1
else:
lt = mid + 1
pos_l = mid
lt, rt = 0, m-1
while lt <= rt:
mid = (lt + rt) // 2
m_num = matrix[mid][-1]
if m_num < target and (mid == m-1 or matrix[mid+1][-1] >= target):
break
elif mid == 0 and m_num >= target:
mid = -1
break
elif m_num >= target:
rt = mid - 1
else:
lt = mid + 1
pos_r = mid + 1
for i in range(pos_r, pos_l+1):
row = matrix[i]
lt, rt = 0, n-1
while lt <= rt:
mid = (lt + rt) // 2
num = row[mid]
if num == target:
return True
elif num > target:
rt = mid - 1
else:
lt = mid + 1
return Falseclass Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
def search(mat, row, col, target):
if col < 0 or row == len(mat):
return False
num = mat[row][col]
if num == target:
return True
elif num > target:
return search(mat, row, col-1, target)
else:
return search(mat, row+1, col, target)
if not len(matrix) or not len(matrix[0]):
return False
return search(matrix, 0, len(matrix[0])-1, target)至此，二分查找的问题学习的差不多了~套路及常见问题基本如此。
下一章节将是最重要的树。