if len(diagonals) == 0 or isscalarlike(diagonals[0]):
时间: 2024-06-04 15:11:09 浏览: 6
This line of code checks if the length of the list called "diagonals" is 0 or if the first element of the list is a scalar (a single value, like an integer or float).
If the length of the list is 0, it means there are no diagonals in the matrix and the code will likely proceed to calculate the diagonal elements in some way.
If the first element of the list is a scalar, it means that the diagonals are not given as a list but rather as a single value that applies to all diagonal elements in the matrix. In this case, the code will likely use this scalar value to populate all diagonal elements in the matrix.
相关问题
if len(diagonals) == 0 or isscalarlike(diagonals[0]):解释len() of unsized object
The error message "len() of unsized object" means that the object being passed to the len() function does not have a defined length or size. This usually occurs when the object is not iterable or does not support the len() method.
In this specific code snippet, the error occurs because the variable "diagonals" has not been initialized or has been initialized as an empty object, thus it does not have a defined length. The condition "len(diagonals) == 0" checks if the length of diagonals is equal to zero or if diagonals is an empty object. If either of these conditions is true, the subsequent code block will execute.
1、(1)请编程实现从 n-皇后问题的所有 92 种可行解中筛选出 12 种独立解,而其余的解都 可以由这些独立解利用对称性或旋转而得到。 (2)若 n-皇后问题要求在求得第一个可行解之后算法即终止,请编程实现。 提示:可以用 flag 标志是否已经求得第一个可行解。根据 flag 的值决定是否继续进行本层 调用,还是直接返回上层调用点。
1. 实现从 n-皇后问题的所有 92 种可行解中筛选出 12 种独立解
以下是 Python 代码实现:
```python
def is_valid(board, row, col):
for i in range(row):
if board[i] == col or abs(board[i] - col) == abs(i - row):
return False
return True
def solve_n_queens(n):
def backtrack(row=0, cols=set(), diagonals=set(), anti_diagonals=set()):
nonlocal count
if row == n:
count += 1
# 收集独立解
if len(visited) < 12:
visited.add(tuple(board))
return
for col in range(n):
diagonal = row - col
anti_diagonal = row + col
if col in cols or diagonal in diagonals or anti_diagonal in anti_diagonals:
continue
board.append(col)
cols.add(col)
diagonals.add(diagonal)
anti_diagonals.add(anti_diagonal)
backtrack(row + 1, cols, diagonals, anti_diagonals)
board.pop()
cols.remove(col)
diagonals.remove(diagonal)
anti_diagonals.remove(anti_diagonal)
count = 0
visited = set()
board = []
backtrack()
return visited
```
2. 实现在求得第一个可行解之后算法即终止
以下是 Python 代码实现:
```python
def solve_n_queens(n):
def backtrack(row=0, cols=set(), diagonals=set(), anti_diagonals=set()):
nonlocal count, flag
if row == n:
count += 1
if flag:
return True
return
for col in range(n):
diagonal = row - col
anti_diagonal = row + col
if col in cols or diagonal in diagonals or anti_diagonal in anti_diagonals:
continue
board.append(col)
cols.add(col)
diagonals.add(diagonal)
anti_diagonals.add(anti_diagonal)
if backtrack(row + 1, cols, diagonals, anti_diagonals):
return True
board.pop()
cols.remove(col)
diagonals.remove(diagonal)
anti_diagonals.remove(anti_diagonal)
return False
count = 0
flag = False
board = []
backtrack()
return count
```
在 `backtrack` 函数中加入 `flag` 标志,当第一个可行解被找到时,将 `flag` 设为 `True`,并在 `backtrack` 函数中返回 True,以终止算法。