已知稀疏矩阵A和B,编程实现基于三元组顺序表实现A+B的运算,请生成三个不同的python程序
时间: 2024-02-20 20:59:04 浏览: 149
程序一:
```python
class Triplet:
def __init__(self, row, col, value):
self.row = row
self.col = col
self.value = value
def add_sparse_matrix(A, B):
rows = max(A[i].row for i in range(len(A))) + 1
cols = max(A[i].col for i in range(len(A))) + 1
C = []
for i in range(rows):
for j in range(cols):
a = 0
b = 0
for k in range(len(A)):
if A[k].row == i and A[k].col == j:
a = A[k].value
break
for k in range(len(B)):
if B[k].row == i and B[k].col == j:
b = B[k].value
break
if a + b != 0:
C.append(Triplet(i, j, a + b))
return C
```
程序二:
```python
class Triplet:
def __init__(self, row, col, value):
self.row = row
self.col = col
self.value = value
def add_sparse_matrix(A, B):
rows = max(A, key=lambda x: x.row).row + 1
cols = max(A, key=lambda x: x.col).col + 1
C = []
i = j = 0
while i < len(A) and j < len(B):
if A[i].row < B[j].row or (A[i].row == B[j].row and A[i].col < B[j].col):
C.append(A[i])
i += 1
elif A[i].row == B[j].row and A[i].col == B[j].col:
if A[i].value + B[j].value != 0:
C.append(Triplet(A[i].row, A[i].col, A[i].value + B[j].value))
i += 1
j += 1
else:
C.append(B[j])
j += 1
while i < len(A):
C.append(A[i])
i += 1
while j < len(B):
C.append(B[j])
j += 1
return C
```
程序三:
```python
class Triplet:
def __init__(self, row, col, value):
self.row = row
self.col = col
self.value = value
def add_sparse_matrix(A, B):
rows = max(A, key=lambda x: x.row).row + 1
cols = max(A, key=lambda x: x.col).col + 1
C = []
for i in range(rows):
for j in range(cols):
a = next((x for x in A if x.row == i and x.col == j), None)
b = next((x for x in B if x.row == i and x.col == j), None)
if a and b:
if a.value + b.value != 0:
C.append(Triplet(i, j, a.value + b.value))
elif a:
C.append(a)
elif b:
C.append(b)
return C
```
三个程序都是基于三元组顺序表实现稀疏矩阵加法的,采用了不同的算法思路,具体实现方式略有差异。
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