python学习|6.2.3 python二分查找算法及LeetCode题目(3)之二维数组 —— Search a 2D Matrix python数据结构与算法|二分查找

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.

【python学习|6.2.3 python二分查找算法及LeetCode题目(3)之二维数组 —— Search a 2D Matrix】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 Falsest = matrix[0][0] ed = matrix[-1][-1] if st > target or ed < target: return Falselt, 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 False

240. 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 Falselt, 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 = midlt, 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 + 1pos_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 False

class 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 Falsereturn 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。