Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements that appear twice in this array.
[4,3,2,7,8,2,3,1]
Output:
[2,3]
首先是笨方法,超时
def findDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
res = []
for i in nums:
num = nums.count(i)
if num > 1 and i not in res:
res.append(i)
return res
网上看的解法,其实就是用-1去标记已出现过的元素
def findDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
res = []
for i in nums:
if nums[abs(i)-1]<0:
res.append(abs(i))
else:
nums[abs(i)-1] *= -1
return res
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