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 + 1Run Time: 72 ms
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