我正在解決問題。您必須保留最后一個串列，但之前的串列會再次列印出來。
def merge(xs,ys):
# xs, ys, ss = xs, ys, []
xs, ys, ss = xs[:], ys[:], []
while xs!=[] and ys!=[]:
if xs[0] <= ys[0]:
ss.append(xs[0])
xs.remove(xs[0])
else:
ss.append(ys[0])
ys.remove(ys[0])
ss.extend(xs)
ss.extend(ys)
return ss
accumulator = []
remain = []
def merge2R(xss):
if len(xss)% 2 != 0 :
OExcept = len(xss)-1
remain.append((xss[OExcept]))
xss.remove(xss[OExcept])
if xss != []:
accumulator.append(merge(xss[0],xss[1]))
xss.remove(xss[0])
xss.remove(xss[0])
return merge2R(xss)
else:
return accumulator remain
結果是這樣的。我該如何解決？
>>> merge2R([[2],[1,3],[4,6,7],[5,8],[9]])
[[1, 2, 3], [4, 5, 6, 7, 8], [1, 2, 3], [4, 5, 6, 7, 8], [9]]
期望的結果值：
>>> merge2R([[2],[1,3],[4,6,7],[5,8]])
[[1,2,3], [4,5,6,7,8]]
>>> merge2R([[2],[1,3],[4,6,7],[5,8],[9]])
[[1,2,3], [4,5,6,7,8], [9]]
uj5u.com熱心網友回復：
您的代碼正在運行，您只需要重置accumulator并remain
def merge2R(xss):
# Declare use of globals
global remain
global accumulator
if len(xss) % 2 != 0:
OExcept = len(xss)-1
remain.append((xss[OExcept]))
xss.remove(xss[OExcept])
if xss != []:
accumulator.append(merge(xss[0], xss[1]))
xss.remove(xss[0])
xss.remove(xss[0])
return merge2R(xss)
else:
x = accumulator remain
# Must reset accumulator and remain
accumulator = []
remain = []
return x
因為您將兩個陣列都初始化為空，所以附加到它們：
# Remain
remain.append((xss[OExcept]))
# Accumulator
accumulator.append(merge(xss[0], xss[1]))
完成這些陣列中的資料后（在函式結束時），您需要丟棄它：
accumulator = []
remain = []
多次呼叫具有相同引數的函式時，不丟棄這些陣列的結果很明顯：
print(merge2R([[2], [1, 3]]))
print(merge2R([[2], [1, 3]]))
print(merge2R([[2], [1, 3]]))
print(merge2R([[2], [1, 3]]))
print(merge2R([[2], [1, 3]]))
[[1, 2, 3]]
[[1, 2, 3], [1, 2, 3]]
[[1, 2, 3], [1, 2, 3], [1, 2, 3]]
[[1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]
[[1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]
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