The solution is using 2 pointers.
Solution:
# T:O(n) S:O(1)
class Solution:
# @param A a list of integers
# @return an integer
# @it's a good solution!
def removeDuplicates(self, A):
if len(A) <= 2: return len(A)
prev = 1; curr = 2
while curr < len(A):
if A[curr] == A[prev] and A[curr] == A[prev - 1]:
curr += 1
else:
prev += 1
A[prev] = A[curr]
curr += 1
return prev + 1
Run Time: 72 ms
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