我正在嘗試編寫一個(gè)遞歸算法，在嵌套數(shù)組中搜索整數(shù)。這就是我目前的代碼，我添加了 print 語句來查看它每次迭代的作用。輸出顯示它應(yīng)該True在某個(gè)時(shí)刻返回，即使它沒有返回。def nestedListContains(lst, n): for i in lst: print(f'i: {i}, n: {n}') if type(i) == list: nestedListContains(i, n) elif int(i) == int(n): return True return Falseprint(nestedListContains([1, [2, [3], 4]], 3)) # Should return Trueprint(nestedListContains([1, [2, [3], 4]], 5)) # Should return False輸出：i: 1, n: 3i: [2, [3], 4], n: 3i: 2, n: 3i: [3], n: 3i: 3, n: 3 # This iteration should return True!i: 4, n: 3Falsei: 1, n: 5i: [2, [3], 4], n: 5i: 2, n: 5i: [3], n: 5i: 3, n: 5i: 4, n: 5False
2 回答
qq_花開花謝_0
TA貢獻(xiàn)1835條經(jīng)驗(yàn) 獲得超7個(gè)贊
當(dāng)您調(diào)用遞歸函數(shù)時(shí)，您缺少 a ，return因?yàn)檫@將是返回的函數(shù)True
def nestedListContains(lst, n):
for i in lst:
print(f'i: {i}, n: {n}')
if type(i) == list:
return nestedListContains(i, n)
elif int(i) == int(n):
return True
return False
print(nestedListContains([1, [2, [3], 4]], 3)) # True
print(nestedListContains([1, [2, [3], 4]], 5)) # False
holdtom
TA貢獻(xiàn)1805條經(jīng)驗(yàn) 獲得超10個(gè)贊
您需要返回調(diào)用的值nestedListContains(i, n)，否則返回的值將被丟棄，并且循環(huán)將繼續(xù)，直到return False到達(dá)：
def nestedListContains(lst, n):
for i in lst:
print(f'i: {i}, n: {n}')
if type(i) == list:
return nestedListContains(i, n)
elif int(i) == int(n):
return True
return False
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