219.Existing duplicate elements II Hash table graphics
Thought
This question was done on the phone at first，直接用的是双Hash table，An element，One depositindex，When a certain element exceeds2Judging whether there are two of them when they areindexSubtission is equal tokCase。So the complexity of time isO(n^3)。
Code：
class Solution:
def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
hash_1={}
hash_2={}
for index, i in enumerate(nums):
if i not in hash_1:
hash_1[i] = 1
hash_2[i] = [index]
else:
hash_1[i] += 1
hash_2[i].append(index)
for i in hash_1:
if hash_1[i] >= 2:
for j in range(len(hash_2[i])):
for m in range(j + 1, len(hash_2[i])):
if abs(hash_2[i][j]-hash_2[i][m]) <= k:
return True
return Falselook@Gongshui SanyeGongshui Sanye的解法后，Only think of，这道题的目的是为了练习双指针在Hash table中的应用，TAThe original words are sorting the meaning：Whether the existence does not exceed k+1k + 1k+1 window，window内有相同元素。
#Sort out the meaning：Whether the existence does not exceed k+1k + 1k+1 window，window内有相同元素。
class Solution:
def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
n = len(nums)
s = set()
for i in range(n):
if i > k:
s.remove(nums[i - k - 1])
if nums[i] in s:
return True
s.add(nums[i])
return False贡献者
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