def find_duplicate(arr): for i in range(1, len(arr)): cur = arr[i] j = i while j > 0 and arr[j - 1] > cur: arr[j] = arr[j - 1] j -= 1 arr[j] = cur if arr[j-1] == cur: arr[j] = cur return f"This is the duplicate {cur}, {arr}" else: arr[j] = cur print("no dupes") return arrprint(find_duplicate([1,-5,3,4,-10,100,250,-325,100]))當(dāng)我在數(shù)組末尾添加 100 時(shí)，它會(huì)被排序，重復(fù)值顯示為 100 并且數(shù)組已正確排序，但是當(dāng)我在末尾添加負(fù)數(shù)（例如 -5）作為重復(fù)值時(shí)，它不會(huì)排序-5 但有些人如何理解這個(gè)數(shù)字是重復(fù)的。我究竟做錯(cuò)了什么？請(qǐng)查看我提到的兩個(gè)數(shù)組作為輸入：arr = [1,-5,3,4,-10,100,250,-325,100] <- 100 結(jié)束arr = [1,-5,3,4,-10,100,250,-325,-5] <- -5 結(jié)束print(find_duplicate([1,-5,3,4,-10,100,250,-325,100]))print(find_duplicate([1,-5,3,4,-10,100,250,-325,-5]))# OUTPUT BELOW:This is the duplicate 100, [-325, -10, -5, 1, 3, 4, 100, 100, 250]This is the duplicate -5, [-5, 1, 3, 4, -10, 100, 250, -325, -5]更新：當(dāng)再次運(yùn)行這個(gè)時(shí)print(find_duplicate([1,-5,3,4,-10,100,250,-325,-5,10]))this is the duplicate -5, [-325, -10, -5, -5, 1, 3, 4, 100, 250, 10]似乎它只是不斷附加數(shù)字，而不是在找到重復(fù)項(xiàng)后對(duì)它們進(jìn)行排序
1 回答
藍(lán)山帝景
TA貢獻(xiàn)1843條經(jīng)驗(yàn) 獲得超7個(gè)贊
你的邏輯是正確的，除了return里面的陳述if
def find_duplicate(arr):
dup = None # set this to none to act as flag
for i in range(1, len(arr)):
cur = arr[i]
j = i
while j > 0 and arr[j - 1] > cur:
arr[j] = arr[j - 1]
j -= 1
arr[j] = cur
if arr[j-1] == cur:
arr[j] = cur
dup = cur
else:
arr[j] = cur
if dup is None: # check if a duplicate was found
print("no dupes")
return arr
else:
return f"This is the duplicate {dup}, {arr}"
print(find_duplicate([1,-5,3,4,-10,100,250,-325,-5]))
print(find_duplicate([1,-5,3,4,-10,100,250,-325,100]))
print(find_duplicate([1,-5,3,4,-10,100,250,-325,-5,10]))
輸出
This is the duplicate -5, [-325, -10, -5, -5, 1, 3, 4, 100, 250]
This is the duplicate 100, [-325, -10, -5, 1, 3, 4, 100, 100, 250]
This is the duplicate -5, [-325, -10, -5, -5, 1, 3, 4, 10, 100, 250]
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